README_EN.md

226. Invert Binary Tree

中文文档

Description

Invert a binary tree.

Example:

Input:

     4

   /   \

  2     7

 / \   / \

1   3 6   9

Output:

     4

   /   \

  7     2

 / \   / \

9   6 3   1

Trivia:

This problem was inspired by this original tweet by Max Howell:

Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so f*** off.

Solutions

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def invertTree(self, root: TreeNode) -> TreeNode:
        if root is None:
            return None
        root.left, root.right = root.right, root.left
        self.invertTree(root.left)
        self.invertTree(root.right)
        return root

Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode invertTree(TreeNode root) {
        if (root == null) return null;
        TreeNode t = root.left;
        root.left = root.right;
        root.right = t;
        invertTree(root.left);
        invertTree(root.right);
        return root;
    }
}

JavaScript

/**
 * Definition for a binary tree node.
 * function TreeNode(val) {
 *     this.val = val;
 *     this.left = this.right = null;
 * }
 */
/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function (root) {
  if (!root) return null;
  [root.left, root.right] = [root.right, root.left];
  invertTree(root.left);
  invertTree(root.right);
  return root;
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func invertTree(root *TreeNode) *TreeNode {
    if (root == nil) {
        return nil
    }
    root.Left, root.Right = root.Right, root.Left
    invertTree(root.Left)
    invertTree(root.Right)
    return root
}

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