Leetcode - Invert Binary Tree

2023年6月24日 175次阅读 来源: Richardo92

My code:

/**
 * 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)
            return root;
        revertTree(root);
        return root;
    }
    
    private void revertTree(TreeNode root) {
        TreeNode temp = root.left;
        root.left = root.right;
        root.right = temp;
        if (root.left != null)
            revertTree(root.left);
        if (root.right != null)
            revertTree(root.right);
    }
}

今天做了三道简单题。。。
这道题木也是最简单的递归思想。
只不过据说这道题木某个大牛面试谷歌的时候没做出来，然后说了如下这么一句话。

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

哈哈。看来技术是和算法数据结构相关的，但也不需要像刷题这样，对他们那么了解，变态的了解。
只不过这么一道题目，其实并不是很难。大牛怎么会做不出来。
还有，现实中，要用到高端算法的情况，好像少之又少。

**
总结： recursive
**

Anyway, Good luck, Richardo!

recursion
My code:

/**
 * 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) {
            return root;
        }
        
        TreeNode left = invertTree(root.left);
        TreeNode right = invertTree(root.right);
        
        root.left = right;
        root.right = left;
        
        return root;
    }
}

iteration
My code:

/**
 * 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) {
            return root;
        }
        
        Queue<TreeNode> q = new LinkedList<TreeNode>();
        q.offer(root);
        
        while (!q.isEmpty()) {
            TreeNode node = q.poll();
            TreeNode left = node.left;
            node.left = node.right;
            node.right = left;
            if (node.left != null) {
                q.offer(node.left);
            }
            if (node.right != null) {
                q.offer(node.right);
            }
        }
        
        return root;
    }
}

iteration 的方法要注意下。这也算是一种新的level order
进去之后不用再循环一次。
再循环一次也差不多。

Anyway, Good luck, Richardo! — 08/28/2016

    原文作者：Richardo92
    原文地址: https://www.jianshu.com/p/a1272a41eea8#comments
    本文转自网络文章，转载此文章仅为分享知识，如有侵权，请联系博主进行删除。
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