N叉树的中序遍历

给定一个包含 N 元树，任务是打印树的中序遍历。

例子：

Input: N = 3 
 

Output: 5 6 2 7 3 1 4
Input: N = 3 
 

Output: 2 5 3 1 4 6 

方法： N 叉树的中序遍历定义为递归地访问除最后一个子节点之外的所有子节点，然后是根节点，最后是最后一个子节点。

• 递归访问第一个孩子。
• 递归访问第二个孩子。
• ……
• 递归访问倒数第二个孩子。
• 打印节点中的数据。
• 递归访问最后一个孩子。
• 重复上述步骤，直到访问完所有节点。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include
using namespace std;
 
// Class for the node of the tree
struct Node
{
    int data;
 
    // List of children
    struct Node **children;
     
    int length;
     
    Node()
    {
        length = 0;
        data = 0;
    }
 
    Node(int n, int data_)
    {
        children = new Node*();
        length = n;
        data = data_;
    }
};
 
// Function to print the inorder traversal
// of the n-ary tree
void inorder(Node *node)
{
    if (node == NULL)
        return;
 
    // Total children count
    int total = node->length;
     
    // All the children except the last
    for (int i = 0; i < total - 1; i++)
        inorder(node->children[i]);
 
    // Print the current node's data
    cout<< node->data << " ";
 
    // Last child
    inorder(node->children[total - 1]);
}
 
// Driver code
int main()
{
 
    /* Create the following tree
          1
        / | \
        2 3 4
        / | \
        5 6 7
    */
    int n = 3;
    Node* root = new Node(n, 1);
    root->children[0] = new Node(n, 2);
    root->children[1] = new Node(n, 3);
    root->children[2] = new Node(n, 4);
    root->children[0]->children[0] = new Node(n, 5);
    root->children[0]->children[1] = new Node(n, 6);
    root->children[0]->children[2] = new Node(n, 7);
 
    inorder(root);
    return 0;
}
 
// This code is Contributed by Arnab Kundu

Java

// Java implementation of the approach
class GFG {
 
    // Class for the node of the tree
    static class Node {
        int data;
 
        // List of children
        Node children[];
 
        Node(int n, int data)
        {
            children = new Node[n];
            this.data = data;
        }
    }
 
    // Function to print the inorder traversal
    // of the n-ary tree
    static void inorder(Node node)
    {
        if (node == null)
            return;
 
        // Total children count
        int total = node.children.length;
        // All the children except the last
        for (int i = 0; i < total - 1; i++)
            inorder(node.children[i]);
 
        // Print the current node's data
        System.out.print("" + node.data + " ");
 
        // Last child
        inorder(node.children[total - 1]);
    }
 
    // Driver code
    public static void main(String[] args)
    {
 
        /* Create the following tree
                   1
                /  |  \
               2   3   4
             / | \
            5  6  7
        */
        int n = 3;
        Node root = new Node(n, 1);
        root.children[0] = new Node(n, 2);
        root.children[1] = new Node(n, 3);
        root.children[2] = new Node(n, 4);
        root.children[0].children[0] = new Node(n, 5);
        root.children[0].children[1] = new Node(n, 6);
        root.children[0].children[2] = new Node(n, 7);
 
        inorder(root);
    }
}

Python3

# Python3 implementation of the approach
class GFG:
     
    # Class for the node of the tree
    class Node:
        def __init__(self,n,data):
            # List of children
            self.children = [None]*n
            self.data = data
     
    # Function to print the inorder traversal
    # of the n-ary tree
    def inorder(self, node):
        if node == None:
            return
         
        # Total children count
        total = len(node.children)
         
        # All the children except the last
        for i in range(total-1):
            self.inorder(node.children[i])
         
        # Print the current node's data
        print(node.data,end=" ")
         
        # Last child
        self.inorder(node.children[total-1])
     
    # Driver code
    def main(self):
        # Create the following tree 
        #          1
        #       /  |  \
        #      2   3   4
        #    / | \
        #   5  6  7
         
        n = 3
        root = self.Node(n, 1)
        root.children[0] = self.Node(n, 2)
        root.children[1] = self.Node(n, 3)
        root.children[2] = self.Node(n, 4)
        root.children[0].children[0] = self.Node(n, 5)
        root.children[0].children[1] = self.Node(n, 6)
        root.children[0].children[2] = self.Node(n, 7)
         
        self.inorder(root)
         
ob = GFG() # Create class object
ob.main() # Call main function
 
# This code is contributed by Shivam Singh

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
    // Class for the node of the tree
    public class Node
    {
        public int data;
 
        // List of children
        public Node []children;
 
        public Node(int n, int data)
        {
            children = new Node[n];
            this.data = data;
        }
    }
 
    // Function to print the inorder traversal
    // of the n-ary tree
    static void inorder(Node node)
    {
        if (node == null)
            return;
 
        // Total children count
        int total = node.children.Length;
         
        // All the children except the last
        for (int i = 0; i < total - 1; i++)
            inorder(node.children[i]);
 
        // Print the current node's data
        Console.Write("" + node.data + " ");
 
        // Last child
        inorder(node.children[total - 1]);
    }
 
    // Driver code
    public static void Main()
    {
 
        /* Create the following tree
                1
                / | \
            2 3 4
            / | \
            5 6 7
        */
        int n = 3;
        Node root = new Node(n, 1);
        root.children[0] = new Node(n, 2);
        root.children[1] = new Node(n, 3);
        root.children[2] = new Node(n, 4);
        root.children[0].children[0] = new Node(n, 5);
        root.children[0].children[1] = new Node(n, 6);
        root.children[0].children[2] = new Node(n, 7);
 
        inorder(root);
    }
}
 
// This code is contributed by AnkitRai01

Javascript

  
// JavaScript implementation of the approach
// Class for the node of the tree
class Node
{
 
    constructor(n, data)
    {
        this.data = data;
        this.children = Array(n);
    }
}
// Function to print the inorder traversal
// of the n-ary tree
function inorder(node)
{
    if (node == null)
        return;
    // Total children count
    var total = node.children.length;
     
    // All the children except the last
    for (var i = 0; i < total - 1; i++)
        inorder(node.children[i]);
    // Print the current node's data
    document.write("" + node.data + " ");
    // Last child
    inorder(node.children[total - 1]);
}
 
// Driver code
/* Create the following tree
        1
        / | \
    2 3 4
    / | \
    5 6 7
*/
var n = 3;
var root = new Node(n, 1);
root.children[0] = new Node(n, 2);
root.children[1] = new Node(n, 3);
root.children[2] = new Node(n, 4);
root.children[0].children[0] = new Node(n, 5);
root.children[0].children[1] = new Node(n, 6);
root.children[0].children[2] = new Node(n, 7);
inorder(root);
 

输出

5 6 2 7 3 1 4 

时间复杂度： O(n)

空间复杂度： O(n)