使用递归的树的 DFS 遍历

给定一棵二叉树，使用递归的 DFS 遍历它。
与只有一种逻辑遍历它们的线性数据结构（数组、链表、队列、堆栈等）不同，树可以以不同的方式遍历。一般来说，有两种广泛使用的遍历树的方法：

• DFS 或深度优先搜索
• BFS 或广度优先搜索

在本文中，已经讨论了使用 DFS 进行遍历。请参阅此帖子以了解广度优先遍历。

深度优先搜索

DFS（深度优先搜索）是用于遍历树或图的技术。这里回溯用于遍历。在这个遍历中，首先访问最深的节点，然后如果该节点的兄弟节点不存在，则回溯到它的父节点。

图与树的 DFS 遍历

在图中，可能存在循环和断开连接。与图不同，树不包含循环并且始终连接。所以树的DFS相对容易。我们可以简单地从一个节点开始，然后遍历它的相邻（或子节点）而不关心循环。如果我们从单个节点（根）开始，并以这种方式遍历，则可以保证我们遍历整个树，因为没有断开连接，

例子：

Tree:

Therefore, the Depth First Traversals of this Tree will be:
(a) Inorder (Left, Root, Right) : 4 2 5 1 3
(b) Preorder (Root, Left, Right) : 1 2 4 5 3
(c) Postorder (Left, Right, Root) : 4 5 2 3 1
 

下面是使用递归的 DFS 遍历树：

1.中序遍历（练习）：

示例：上图的中序遍历是 4 2 5 1 3。

Algorithm Inorder(tree)
   1. Traverse the left subtree, i.e., call Inorder(left-subtree)
   2. Visit the root.
   3. Traverse the right subtree, i.e., call Inorder(right-subtree)

执行：

C++

// C program for different tree traversals
#include 
using namespace std;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node {
    int data;
    struct Node *left, *right;
    Node(int data)
    {
        this->data = data;
        left = right = NULL;
    }
};
 
/* Given a binary tree, print its nodes in inorder*/
void printInorder(struct Node* node)
{
    if (node == NULL)
        return;
 
    /* first recur on left child */
    printInorder(node->left);
 
    /* then print the data of node */
    cout << node->data << " ";
 
    /* now recur on right child */
    printInorder(node->right);
}
 
/* Driver program to test above functions*/
int main()
{
    struct Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->left->right = new Node(5);
 
    cout << "\nInorder traversal of binary tree is \n";
    printInorder(root);
 
    return 0;
}

C

// C program for different tree traversals
#include 
#include 
 
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node {
    int data;
    struct node* left;
    struct node* right;
};
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct node* newNode(int data)
{
    struct node* node = (struct node*)
        malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
 
    return (node);
}
 
/* Given a binary tree, print its nodes in inorder*/
void printInorder(struct node* node)
{
    if (node == NULL)
        return;
 
    /* first recur on left child */
    printInorder(node->left);
 
    /* then print the data of node */
    printf("%d ", node->data);
 
    /* now recur on right child */
    printInorder(node->right);
}
 
/* Driver program to test above functions*/
int main()
{
    struct node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
 
    printf("\nInorder traversal of binary tree is \n");
    printInorder(root);
 
    getchar();
    return 0;
}

Java

// Java program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    int key;
    Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class BinaryTree {
    // Root of Binary Tree
    Node root;
 
    BinaryTree()
    {
        root = null;
    }
 
    /* Given a binary tree, print its nodes in inorder*/
    void printInorder(Node node)
    {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        System.out.print(node.key + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    // Wrappers over above recursive functions
    void printInorder() { printInorder(root); }
 
    // Driver method
    public static void main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
 
        System.out.println("\nInorder traversal of binary tree is ");
        tree.printInorder();
    }
}

Python

# Python program to for tree traversals
 
# A class that represents an individual node in a
# Binary Tree
class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.val = key
 
 
# A function to do inorder tree traversal
def printInorder(root):
 
    if root:
 
        # First recur on left child
        printInorder(root.left)
 
        # then print the data of node
        print(root.val),
 
        # now recur on right child
        printInorder(root.right)
         
 
# Driver code
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
 
print "\nInorder traversal of binary tree is"
printInorder(root)

C#

// C# program for different tree traversals
using System;
     
/* Class containing left and right child of current
node and key value*/
class Node
{
    public int key;
    public Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
public class BinaryTree
{
    // Root of Binary Tree
    Node root;
 
    BinaryTree()
    {
        root = null;
    }
 
    /* Given a binary tree, print its nodes in inorder*/
    void printInorder(Node node)
    {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        Console.Write(node.key + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    // Wrappers over above recursive functions
    void printInorder()
    {
        printInorder(root);
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
 
        Console.WriteLine("\nInorder traversal of binary tree is ");
        tree.printInorder();
    }
}
 
// This code is contributed by PrinciRaj1992

Javascript

// javascript program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    constructor(val) {
        this.key = val;
        this.left = null;
        this.right = null;
    }
}
     
 
    /* Given a binary tree, print its nodes in inorder */
     function printInorder(node) {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        document.write(node.key + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    // Driver method
 
        var root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
 
        document.write("<br/>Inorder traversal of binary tree is <br/>");
        printInorder(root);
 
// This code is contributed by umadevi9616

输出：

Inorder traversal of binary tree is 
4 2 5 1 3

顺序的用途：
在二叉搜索树 (BST) 的情况下，中序遍历以非递减顺序给出节点。为了以非递增顺序获取 BST 的节点，可以使用 Inorder traversal 的变体，其中 Inorder traversal s reversed。

2. 前序遍历（练习）：

示例：上图的前序遍历是 1 2 4 5 3。

Algorithm Preorder(tree)
   1. Visit the root.
   2. Traverse the left subtree, i.e., call Preorder(left-subtree)
   3. Traverse the right subtree, i.e., call Preorder(right-subtree)

执行：

C++

// C program for different tree traversals
#include 
using namespace std;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node {
    int data;
    struct Node *left, *right;
    Node(int data)
    {
        this->data = data;
        left = right = NULL;
    }
};
 
/* Given a binary tree, print its nodes in preorder*/
void printPreorder(struct Node* node)
{
    if (node == NULL)
        return;
 
    /* first print data of node */
    cout << node->data << " ";
 
    /* then recur on left subtree */
    printPreorder(node->left);
 
    /* now recur on right subtree */
    printPreorder(node->right);
}
 
/* Driver program to test above functions*/
int main()
{
    struct Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->left->right = new Node(5);
 
    cout << "\nPreorder traversal of binary tree is \n";
    printPreorder(root);
 
    return 0;
}

C

// C program for different tree traversals
#include 
#include 
 
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node {
    int data;
    struct node* left;
    struct node* right;
};
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct node* newNode(int data)
{
    struct node* node = (struct node*)
        malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
 
    return (node);
}
 
/* Given a binary tree, print its nodes in preorder*/
void printPreorder(struct node* node)
{
    if (node == NULL)
        return;
 
    /* first print data of node */
    printf("%d ", node->data);
 
    /* then recur on left subtree */
    printPreorder(node->left);
 
    /* now recur on right subtree */
    printPreorder(node->right);
}
 
/* Driver program to test above functions*/
int main()
{
    struct node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
 
    printf("\nPreorder traversal of binary tree is \n");
    printPreorder(root);
 
    getchar();
    return 0;
}

Java

// Java program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    int key;
    Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class BinaryTree {
    // Root of Binary Tree
    Node root;
 
    BinaryTree()
    {
        root = null;
    }
 
    /* Given a binary tree, print its nodes in preorder*/
    void printPreorder(Node node)
    {
        if (node == null)
            return;
 
        /* first print data of node */
        System.out.print(node.key + " ");
 
        /* then recur on left subtree */
        printPreorder(node.left);
 
        /* now recur on right subtree */
        printPreorder(node.right);
    }
 
    // Wrappers over above recursive functions
    void printPreorder() { printPreorder(root); }
 
    // Driver method
    public static void main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
 
        System.out.println("Preorder traversal of binary tree is ");
        tree.printPreorder();
    }
}

Python

# Python program to for tree traversals
 
# A class that represents an individual node in a
# Binary Tree
class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.val = key
 
# A function to do preorder tree traversal
def printPreorder(root):
 
    if root:
 
        # First print the data of node
        print(root.val),
 
        # Then recur on left child
        printPreorder(root.left)
 
        # Finally recur on right child
        printPreorder(root.right)
 
 
# Driver code
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
print "Preorder traversal of binary tree is"
printPreorder(root)

C＃

// C# program for different tree traversals
using System;
 
/* Class containing left and right child of current
node and key value*/
public class Node
{
    public int key;
    public Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
public class BinaryTree
{
    // Root of Binary Tree
    Node root;
 
    BinaryTree()
    {
        root = null;
    }
 
    /* Given a binary tree, print its nodes in preorder*/
    void printPreorder(Node node)
    {
        if (node == null)
            return;
 
        /* first print data of node */
        Console.Write(node.key + " ");
 
        /* then recur on left subtree */
        printPreorder(node.left);
 
        /* now recur on right subtree */
        printPreorder(node.right);
    }
 
    // Wrappers over above recursive functions
    void printPreorder() { printPreorder(root); }
 
    // Driver method
    public static void Main()
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
 
        Console.WriteLine("Preorder traversal of binary tree is ");
        tree.printPreorder();
    }
}
 
/* This code contributed by PrinciRaj1992 */

输出：

Preorder traversal of binary tree is 
1 2 4 5 3

预购的用途：
前序遍历用于创建树的副本。前序遍历也用于获取表达式树的前缀表达式。请参阅 http://en.wikipedia.org/wiki/Polish_notation 了解前缀表达式为何有用。

3.后序遍历（练习）：

示例：上述给定树的后序遍历是 4 5 2 3 1。

Algorithm Postorder(tree)
   1. Traverse the left subtree, i.e., call Postorder(left-subtree)
   2. Traverse the right subtree, i.e., call Postorder(right-subtree)
   3. Visit the root.

执行：

C++

// C program for different tree traversals
#include 
using namespace std;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node {
    int data;
    struct Node *left, *right;
    Node(int data)
    {
        this->data = data;
        left = right = NULL;
    }
};
 
/* Given a binary tree, print its nodes according to the
"bottom-up" postorder traversal. */
void printPostorder(struct Node* node)
{
    if (node == NULL)
        return;
 
    // first recur on left subtree
    printPostorder(node->left);
 
    // then recur on right subtree
    printPostorder(node->right);
 
    // now deal with the node
    cout << node->data << " ";
}
 
/* Driver program to test above functions*/
int main()
{
    struct Node* root = new Node(1);
    root->left = new Node(2);
    root->right = new Node(3);
    root->left->left = new Node(4);
    root->left->right = new Node(5);
 
    cout << "\nPostorder traversal of binary tree is \n";
    printPostorder(root);
 
    return 0;
}

C

// C program for different tree traversals
#include 
#include 
 
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node {
    int data;
    struct node* left;
    struct node* right;
};
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct node* newNode(int data)
{
    struct node* node = (struct node*)
        malloc(sizeof(struct node));
    node->data = data;
    node->left = NULL;
    node->right = NULL;
 
    return (node);
}
 
/* Given a binary tree, print its nodes according to the
  "bottom-up" postorder traversal. */
void printPostorder(struct node* node)
{
    if (node == NULL)
        return;
 
    // first recur on left subtree
    printPostorder(node->left);
 
    // then recur on right subtree
    printPostorder(node->right);
 
    // now deal with the node
    printf("%d ", node->data);
}
 
/* Driver program to test above functions*/
int main()
{
    struct node* root = newNode(1);
    root->left = newNode(2);
    root->right = newNode(3);
    root->left->left = newNode(4);
    root->left->right = newNode(5);
 
    printf("\nPostorder traversal of binary tree is \n");
    printPostorder(root);
 
    getchar();
    return 0;
}

Java

// Java program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    int key;
    Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class BinaryTree {
    // Root of Binary Tree
    Node root;
 
    BinaryTree()
    {
        root = null;
    }
 
    /* Given a binary tree, print its nodes according to the
      "bottom-up" postorder traversal. */
    void printPostorder(Node node)
    {
        if (node == null)
            return;
 
        // first recur on left subtree
        printPostorder(node.left);
 
        // then recur on right subtree
        printPostorder(node.right);
 
        // now deal with the node
        System.out.print(node.key + " ");
    }
 
    // Wrappers over above recursive functions
    void printPostorder() { printPostorder(root); }
 
    // Driver method
    public static void main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
 
        System.out.println("\nPostorder traversal of binary tree is ");
        tree.printPostorder();
    }
}

Python

# Python program to for tree traversals
 
# A class that represents an individual node in a
# Binary Tree
class Node:
    def __init__(self, key):
        self.left = None
        self.right = None
        self.val = key
 
         
# A function to do postorder tree traversal
def printPostorder(root):
 
    if root:
 
        # First recur on left child
        printPostorder(root.left)
 
        # the recur on right child
        printPostorder(root.right)
 
        # now print the data of node
        print(root.val),
 
 
# Driver code
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
 
print "\nPostorder traversal of binary tree is"
printPostorder(root)

C＃

// C# program for different tree traversals
using System;
 
/* Class containing left and right child of current
node and key value*/
public class Node
{
    public int key;
    public Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
public class BinaryTree
{
    // Root of Binary Tree
    Node root;
 
    BinaryTree()
    {
        root = null;
    }
 
    /* Given a binary tree, print its nodes according to the
    "bottom-up" postorder traversal. */
    void printPostorder(Node node)
    {
        if (node == null)
            return;
 
        // first recur on left subtree
        printPostorder(node.left);
 
        // then recur on right subtree
        printPostorder(node.right);
 
        // now deal with the node
        Console.Write(node.key + " ");
    }
 
    // Wrappers over above recursive functions
    void printPostorder() { printPostorder(root); }
 
    // Driver code
    public static void Main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
 
        Console.WriteLine("\nPostorder traversal of binary tree is ");
        tree.printPostorder();
    }
}
 
// This code contributed by Rajput-Ji

Javascript

// javascript program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node {
    constructor(item) {
        this.key = item;
        this.left = this.right = null;
    }
}
 
var root;
 
    /*
     * Given a binary tree, print its nodes according to the "bottom-up" postorder
     * traversal.
     */
    function printPostorder(node) {
        if (node == null)
            return;
 
        // first recur on left subtree
        printPostorder(node.left);
 
        // then recur on right subtree
        printPostorder(node.right);
 
        // now deal with the node
        document.write(node.key + " ");
    }
 
     
    // Driver method
 
        root = new Node(1);
        root.left = new Node(2);
        root.right = new Node(3);
        root.left.left = new Node(4);
        root.left.right = new Node(5);
 
        document.write("\nPostorder traversal of binary tree is<br/> ");
        printPostorder(root);
// This code contributed by umadevi9616

输出：

Postorder traversal of binary tree is 
4 5 2 3 1

后序的用途：
后序遍历用于删除树。有关详细信息，请参阅删除树的问题。后序遍历对于获取表达式树的后缀表达式也很有用。有关后缀表达式的用法，请参阅 http://en.wikipedia.org/wiki/Reverse_Polish_notation。

使用 DFS 实现所有遍历

C++

// C program for different tree traversals
#include 
using namespace std;
 
/* A binary tree node has data, pointer to left child
and a pointer to right child */
struct Node
{
    int data;
    struct Node* left, *right;
    Node(int data)
    {
        this->data = data;
        left = right = NULL;
    }
};
 
/* Given a binary tree, print its nodes according to the
"bottom-up" postorder traversal. */
void printPostorder(struct Node* node)
{
    if (node == NULL)
        return;
 
    // first recur on left subtree
    printPostorder(node->left);
 
    // then recur on right subtree
    printPostorder(node->right);
 
    // now deal with the node
    cout << node->data << " ";
}
 
/* Given a binary tree, print its nodes in inorder*/
void printInorder(struct Node* node)
{
    if (node == NULL)
        return;
 
    /* first recur on left child */
    printInorder(node->left);
 
    /* then print the data of node */
    cout << node->data << " ";
 
    /* now recur on right child */
    printInorder(node->right);
}
 
/* Given a binary tree, print its nodes in preorder*/
void printPreorder(struct Node* node)
{
    if (node == NULL)
        return;
 
    /* first print data of node */
    cout << node->data << " ";
 
    /* then recur on left subtree */
    printPreorder(node->left);
 
    /* now recur on right subtree */
    printPreorder(node->right);
}
 
/* Driver program to test above functions*/
int main()
{
    struct Node *root = new Node(1);
    root->left             = new Node(2);
    root->right         = new Node(3);
    root->left->left     = new Node(4);
    root->left->right = new Node(5);
 
    cout << "\nPreorder traversal of binary tree is \n";
    printPreorder(root);
 
    cout << "\nInorder traversal of binary tree is \n";
    printInorder(root);
 
    cout << "\nPostorder traversal of binary tree is \n";
    printPostorder(root);
 
    return 0;
}

C

// C program for different tree traversals
#include 
#include 
 
/* A binary tree node has data, pointer to left child
   and a pointer to right child */
struct node
{
     int data;
     struct node* left;
     struct node* right;
};
 
/* Helper function that allocates a new node with the
   given data and NULL left and right pointers. */
struct node* newNode(int data)
{
     struct node* node = (struct node*)
                                  malloc(sizeof(struct node));
     node->data = data;
     node->left = NULL;
     node->right = NULL;
 
     return(node);
}
 
/* Given a binary tree, print its nodes according to the
  "bottom-up" postorder traversal. */
void printPostorder(struct node* node)
{
     if (node == NULL)
        return;
 
     // first recur on left subtree
     printPostorder(node->left);
 
     // then recur on right subtree
     printPostorder(node->right);
 
     // now deal with the node
     printf("%d ", node->data);
}
 
/* Given a binary tree, print its nodes in inorder*/
void printInorder(struct node* node)
{
     if (node == NULL)
          return;
 
     /* first recur on left child */
     printInorder(node->left);
 
     /* then print the data of node */
     printf("%d ", node->data); 
 
     /* now recur on right child */
     printInorder(node->right);
}
 
/* Given a binary tree, print its nodes in preorder*/
void printPreorder(struct node* node)
{
     if (node == NULL)
          return;
 
     /* first print data of node */
     printf("%d ", node->data); 
 
     /* then recur on left subtree */
     printPreorder(node->left); 
 
     /* now recur on right subtree */
     printPreorder(node->right);
}   
 
/* Driver program to test above functions*/
int main()
{
     struct node *root  = newNode(1);
     root->left             = newNode(2);
     root->right           = newNode(3);
     root->left->left     = newNode(4);
     root->left->right   = newNode(5);
 
     printf("\nPreorder traversal of binary tree is \n");
     printPreorder(root);
 
     printf("\nInorder traversal of binary tree is \n");
     printInorder(root); 
 
     printf("\nPostorder traversal of binary tree is \n");
     printPostorder(root);
 
     getchar();
     return 0;
}

Java

// Java program for different tree traversals
 
/* Class containing left and right child of current
   node and key value*/
class Node
{
    int key;
    Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
class BinaryTree
{
    // Root of Binary Tree
    Node root;
 
    BinaryTree()
    {
        root = null;
    }
 
    /* Given a binary tree, print its nodes according to the
      "bottom-up" postorder traversal. */
    void printPostorder(Node node)
    {
        if (node == null)
            return;
 
        // first recur on left subtree
        printPostorder(node.left);
 
        // then recur on right subtree
        printPostorder(node.right);
 
        // now deal with the node
        System.out.print(node.key + " ");
    }
 
    /* Given a binary tree, print its nodes in inorder*/
    void printInorder(Node node)
    {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        System.out.print(node.key + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    /* Given a binary tree, print its nodes in preorder*/
    void printPreorder(Node node)
    {
        if (node == null)
            return;
 
        /* first print data of node */
        System.out.print(node.key + " ");
 
        /* then recur on left subtree */
        printPreorder(node.left);
 
        /* now recur on right subtree */
        printPreorder(node.right);
    }
 
    // Wrappers over above recursive functions
    void printPostorder()  {     printPostorder(root);  }
    void printInorder()    {     printInorder(root);   }
    void printPreorder()   {     printPreorder(root);  }
 
    // Driver method
    public static void main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
 
        System.out.println("Preorder traversal of binary tree is ");
        tree.printPreorder();
 
        System.out.println("\nInorder traversal of binary tree is ");
        tree.printInorder();
 
        System.out.println("\nPostorder traversal of binary tree is ");
        tree.printPostorder();
    }
}

Python

# Python program to for tree traversals
 
# A class that represents an individual node in a
# Binary Tree
class Node:
    def __init__(self,key):
        self.left = None
        self.right = None
        self.val = key
 
 
# A function to do inorder tree traversal
def printInorder(root):
 
    if root:
 
        # First recur on left child
        printInorder(root.left)
 
        # then print the data of node
        print(root.val),
 
        # now recur on right child
        printInorder(root.right)
 
 
 
# A function to do postorder tree traversal
def printPostorder(root):
 
    if root:
 
        # First recur on left child
        printPostorder(root.left)
 
        # the recur on right child
        printPostorder(root.right)
 
        # now print the data of node
        print(root.val),
 
 
# A function to do preorder tree traversal
def printPreorder(root):
 
    if root:
 
        # First print the data of node
        print(root.val),
 
        # Then recur on left child
        printPreorder(root.left)
 
        # Finally recur on right child
        printPreorder(root.right)
 
 
# Driver code
root = Node(1)
root.left      = Node(2)
root.right     = Node(3)
root.left.left  = Node(4)
root.left.right  = Node(5)
print "Preorder traversal of binary tree is"
printPreorder(root)
 
print "\nInorder traversal of binary tree is"
printInorder(root)
 
print "\nPostorder traversal of binary tree is"
printPostorder(root)

C＃

// C# program for different Console.Writetree traversals
 
using System;
 
/* Class containing left and right child of current
node and key value*/
public class Node
{
    public int key;
    public Node left, right;
 
    public Node(int item)
    {
        key = item;
        left = right = null;
    }
}
 
public class BinaryTree
{
    // Root of Binary Tree
    Node root;
 
    BinaryTree()
    {
        root = null;
    }
 
    /* Given a binary tree, print
    its nodes according to the
    "bottom-up" postorder traversal. */
    void printPostorder(Node node)
    {
        if (node == null)
            return;
 
        // first recur on left subtree
        printPostorder(node.left);
 
        // then recur on right subtree
        printPostorder(node.right);
 
        // now deal with the node
        Console.Write(node.key + " ");
    }
 
    /* Given a binary tree, print its nodes in inorder*/
    void printInorder(Node node)
    {
        if (node == null)
            return;
 
        /* first recur on left child */
        printInorder(node.left);
 
        /* then print the data of node */
        Console.Write(node.key + " ");
 
        /* now recur on right child */
        printInorder(node.right);
    }
 
    /* Given a binary tree, print its nodes in preorder*/
    void printPreorder(Node node)
    {
        if (node == null)
            return;
 
        /* first print data of node */
        Console.Write(node.key + " ");
 
        /* then recur on left subtree */
        printPreorder(node.left);
 
        /* now recur on right subtree */
        printPreorder(node.right);
    }
 
    // Wrappers over above recursive functions
    void printPostorder() { printPostorder(root); }
    void printInorder() { printInorder(root); }
    void printPreorder() { printPreorder(root); }
 
    // Driver code
    public static void Main(String[] args)
    {
        BinaryTree tree = new BinaryTree();
        tree.root = new Node(1);
        tree.root.left = new Node(2);
        tree.root.right = new Node(3);
        tree.root.left.left = new Node(4);
        tree.root.left.right = new Node(5);
 
        Console.WriteLine("Preorder traversal of binary tree is ");
        tree.printPreorder();
 
        Console.WriteLine("\nInorder traversal of binary tree is ");
        tree.printInorder();
 
        Console.WriteLine("\nPostorder traversal of binary tree is ");
        tree.printPostorder();
    }
}
 
// This code has been contributed by 29AjayKumar

输出：

Preorder traversal of binary tree is
1 2 4 5 3 
Inorder traversal of binary tree is
4 2 5 1 3 
Postorder traversal of binary tree is
4 5 2 3 1

时间复杂度： O(n)
辅助空间：如果我们不考虑函数调用的堆栈大小，则 O(1) 否则 O(n)。