[LeetCode] 226. Invert Binary Tree

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.

翻转二叉树。也是一道不会做，会写homebrew也枉然的题。题干即是题意。例子如下，即层层遍历，把左右子树交换。

两种思路，分别是层序遍历（BFS）和深度遍历（DFS）。

DFS

时间O(n)

空间O(n)

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function(root) {
    if (root === null) return root;
    let left = invertTree(root.left);
    let right = invertTree(root.right);
    root.left = right;
    root.right = left;
    return root;
};

Java实现

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

BFS

时间O(n)

空间O(n)

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {TreeNode}
 */
var invertTree = function(root) {
    if (!root) return root;
    let queue = [root];
    while (queue.length) {
        let current = queue.shift();
        if (current === null) continue;
        swap(current);
        queue.push(current.left);
        queue.push(current.right);
    }
    return root;
};

var swap = tree => {
    let temp = tree.left;
    tree.left = tree.right;
    tree.right = temp;
    return tree;
};

Java实现

class Solution {
    public TreeNode invertTree(TreeNode root) {
        // corner case
        if (root == null) {
            return root;
        }

        // normal case
        Queue<TreeNode> queue = new LinkedList<>();
        queue.offer(root);
        while (!queue.isEmpty()) {
            TreeNode cur = queue.poll();
            TreeNode temp = cur.left;
            cur.left = cur.right;
            cur.right = temp;
            if (cur.left != null) {
                queue.offer(cur.left);
            }
            if (cur.right != null) {
                queue.offer(cur.right);
            }
        }
        return root;
    }
}

LeetCode 题目总结