[LeetCode] Invert Binary Tree

Invert a binary tree.

     4
   /   \   2     7
 / \   / \ 1   3 6   9

to

     4
   /   \   7     2
 / \   / \ 9   6 3   1

解题思路

递归

实现代码

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:
    TreeNode* 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; } * } */
public class Solution {
    public TreeNode 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;
    }
}

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):
        if root != None:
            root.left, root.right = root.right, root.left
            self.invertTree(root.left)
            self.invertTree(root.right)
        return root