无递归的中序树遍历
使用堆栈是无需递归即可遍历树的明显方法。下面是一个使用栈遍历二叉树的算法。有关算法的逐步执行，请参见this。
1) Create an empty stack S.
2) Initialize current node as root
3) Push the current node to S and set current = current->left until current is NULL
4) If current is NULL and stack is not empty then
a) Pop the top item from stack.
b) Print the popped item, set current = popped_item->right
c) Go to step 3.
5) If current is NULL and stack is empty then we are done.让我们以下面的树为例
1
/ \
2 3
/ \
4 5
Step 1 Creates an empty stack: S = NULL
Step 2 sets current as address of root: current -> 1
Step 3 Pushes the current node and set current = current->left
until current is NULL
current -> 1
push 1: Stack S -> 1
current -> 2
push 2: Stack S -> 2, 1
current -> 4
push 4: Stack S -> 4, 2, 1
current = NULL
Step 4 pops from S
a) Pop 4: Stack S -> 2, 1
b) print "4"
c) current = NULL /*right of 4 */ and go to step 3
Since current is NULL step 3 doesn't do anything.
Step 4 pops again.
a) Pop 2: Stack S -> 1
b) print "2"
c) current -> 5/*right of 2 */ and go to step 3
Step 3 pushes 5 to stack and makes current NULL
Stack S -> 5, 1
current = NULL
Step 4 pops from S
a) Pop 5: Stack S -> 1
b) print "5"
c) current = NULL /*right of 5 */ and go to step 3
Since current is NULL step 3 doesn't do anything
Step 4 pops again.
a) Pop 1: Stack S -> NULL
b) print "1"
c) current -> 3 /*right of 1 */
Step 3 pushes 3 to stack and makes current NULL
Stack S -> 3
current = NULL
Step 4 pops from S
a) Pop 3: Stack S -> NULL
b) print "3"
c) current = NULL /*right of 3 */
Traversal is done now as stack S is empty and current is NULL.C++
// C++ program to print inorder traversal
// using stack.
#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;
struct Node* right;
Node (int data)
{
this->data = data;
left = right = NULL;
}
};
/* Iterative function for inorder tree
traversal */
void inOrder(struct Node *root)
{
stack s;
Node *curr = root;
while (curr != NULL || s.empty() == false)
{
/* Reach the left most Node of the
curr Node */
while (curr != NULL)
{
/* place pointer to a tree node on
the stack before traversing
the node's left subtree */
s.push(curr);
curr = curr->left;
}
/* Current must be NULL at this point */
curr = s.top();
s.pop();
cout << curr->data << " ";
/* we have visited the node and its
left subtree. Now, it's right
subtree's turn */
curr = curr->right;
} /* end of while */
}
/* Driver program to test above functions*/
int main()
{
/* Constructed binary tree is
1
/ \
2 3
/ \
4 5
*/
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);
inOrder(root);
return 0;
} C
#include
#include
#define bool int
/* A binary tree tNode has data, pointer to left child
and a pointer to right child */
struct tNode
{
int data;
struct tNode* left;
struct tNode* right;
};
/* Structure of a stack node. Linked List implementation is used for
stack. A stack node contains a pointer to tree node and a pointer to
next stack node */
struct sNode
{
struct tNode *t;
struct sNode *next;
};
/* Stack related functions */
void push(struct sNode** top_ref, struct tNode *t);
struct tNode *pop(struct sNode** top_ref);
bool isEmpty(struct sNode *top);
/* Iterative function for inorder tree traversal */
void inOrder(struct tNode *root)
{
/* set current to root of binary tree */
struct tNode *current = root;
struct sNode *s = NULL; /* Initialize stack s */
bool done = 0;
while (!done)
{
/* Reach the left most tNode of the current tNode */
if(current != NULL)
{
/* place pointer to a tree node on the stack before traversing
the node's left subtree */
push(&s, current);
current = current->left;
}
/* backtrack from the empty subtree and visit the tNode
at the top of the stack; however, if the stack is empty,
you are done */
else
{
if (!isEmpty(s))
{
current = pop(&s);
printf("%d ", current->data);
/* we have visited the node and its left subtree.
Now, it's right subtree's turn */
current = current->right;
}
else
done = 1;
}
} /* end of while */
}
/* UTILITY FUNCTIONS */
/* Function to push an item to sNode*/
void push(struct sNode** top_ref, struct tNode *t)
{
/* allocate tNode */
struct sNode* new_tNode =
(struct sNode*) malloc(sizeof(struct sNode));
if(new_tNode == NULL)
{
printf("Stack Overflow \n");
getchar();
exit(0);
}
/* put in the data */
new_tNode->t = t;
/* link the old list off the new tNode */
new_tNode->next = (*top_ref);
/* move the head to point to the new tNode */
(*top_ref) = new_tNode;
}
/* The function returns true if stack is empty, otherwise false */
bool isEmpty(struct sNode *top)
{
return (top == NULL)? 1 : 0;
}
/* Function to pop an item from stack*/
struct tNode *pop(struct sNode** top_ref)
{
struct tNode *res;
struct sNode *top;
/*If sNode is empty then error */
if(isEmpty(*top_ref))
{
printf("Stack Underflow \n");
getchar();
exit(0);
}
else
{
top = *top_ref;
res = top->t;
*top_ref = top->next;
free(top);
return res;
}
}
/* Helper function that allocates a new tNode with the
given data and NULL left and right pointers. */
struct tNode* newtNode(int data)
{
struct tNode* tNode = (struct tNode*)
malloc(sizeof(struct tNode));
tNode->data = data;
tNode->left = NULL;
tNode->right = NULL;
return(tNode);
}
/* Driver program to test above functions*/
int main()
{
/* Constructed binary tree is
1
/ \
2 3
/ \
4 5
*/
struct tNode *root = newtNode(1);
root->left = newtNode(2);
root->right = newtNode(3);
root->left->left = newtNode(4);
root->left->right = newtNode(5);
inOrder(root);
getchar();
return 0;
} Java
// non-recursive java program for inorder traversal
import java.util.Stack;
/* Class containing left and right child of
current node and key value*/
class Node
{
int data;
Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
/* Class to print the inorder traversal */
class BinaryTree
{
Node root;
void inorder()
{
if (root == null)
return;
Stack s = new Stack();
Node curr = root;
// traverse the tree
while (curr != null || s.size() > 0)
{
/* Reach the left most Node of the
curr Node */
while (curr != null)
{
/* place pointer to a tree node on
the stack before traversing
the node's left subtree */
s.push(curr);
curr = curr.left;
}
/* Current must be NULL at this point */
curr = s.pop();
System.out.print(curr.data + " ");
/* we have visited the node and its
left subtree. Now, it's right
subtree's turn */
curr = curr.right;
}
}
public static void main(String args[])
{
/* creating a binary tree and entering
the nodes */
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);
tree.inorder();
}
} Python3
# Python program to do inorder traversal without recursion
# A binary tree node
class Node:
# Constructor to create a new node
def __init__(self, data):
self.data = data
self.left = None
self.right = None
# Iterative function for inorder tree traversal
def inOrder(root):
# Set current to root of binary tree
current = root
stack = [] # initialize stack
while True:
# Reach the left most Node of the current Node
if current is not None:
# Place pointer to a tree node on the stack
# before traversing the node's left subtree
stack.append(current)
current = current.left
# BackTrack from the empty subtree and visit the Node
# at the top of the stack; however, if the stack is
# empty you are done
elif(stack):
current = stack.pop()
print(current.data, end=" ") # Python 3 printing
# We have visited the node and its left
# subtree. Now, it's right subtree's turn
current = current.right
else:
break
print()
# Driver program to test above function
""" Constructed binary tree is
1
/ \
2 3
/ \
4 5 """
root = Node(1)
root.left = Node(2)
root.right = Node(3)
root.left.left = Node(4)
root.left.right = Node(5)
inOrder(root)
# This code is contributed by Nikhil Kumar Singh(nickzuck_007)C#
// Non-recursive C# program for inorder traversal
using System;
using System.Collections.Generic;
/* Class containing left and right child of
current node and key value*/
public class Node
{
public int data;
public Node left, right;
public Node(int item)
{
data = item;
left = right = null;
}
}
/* Class to print the inorder traversal */
public class BinaryTree
{
public Node root;
public virtual void inorder()
{
if (root == null)
{
return;
}
Stack s = new Stack();
Node curr = root;
// traverse the tree
while (curr != null || s.Count > 0)
{
/* Reach the left most Node of the
curr Node */
while (curr != null)
{
/* place pointer to a tree node on
the stack before traversing
the node's left subtree */
s.Push(curr);
curr = curr.left;
}
/* Current must be NULL at this point */
curr = s.Pop();
Console.Write(curr.data + " ");
/* we have visited the node and its
left subtree. Now, it's right
subtree's turn */
curr = curr.right;
}
}
public static void Main(string[] args)
{
/* creating a binary tree and entering
the nodes */
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);
tree.inorder();
}
}
// This code is contributed by Shrikant13 Javascript
输出：
4 2 5 1 3