[Leetcode] Invert Binary Tree

Invert a binary tree.

     4   /     2     \   1   3 6   9

To

     4   /     7     \   9   6 3   1

Thinking of solving problems

Recursive

Implementation code

C++:

// Runtime: 3 ms/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    invertTree(TreeNode* root) {        if (root != NULL)        {            TreeNode *temp = root->left;            root->left = root->right;            root->right = temp;            invertTree(root->left);            invertTree(root->right);        }        return root;    }};

Java:

// Runtime: 238 ms/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */publicclass Solution {    publicinvertTree(TreeNode root) {        ifnull){            TreeNode temp = root.left;            root.left = root.right;            root.right = temp;            invertTree(root.left);            invertTree(root.right);        }        return root;    }}

Python:

# runtime:52 MS# Definition for a binary tree node.# class TreeNode:# def __init__ (self, x):# self.val = x# self.left = None# self.right = None class solution:    # @param {TreeNode} root    # @return {TreeNode}     def inverttree(self, root):        ifRoot! =None: root.left, root.right = Root.right, Root.left self.inverttree (root.left) self.inverttre E (root.right)returnRoot

[Leetcode] Invert Binary Tree