226. Invert Binary Tree

 

Invert a binary tree.

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

to

     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 fuck off.

 

这道题让我们翻转二叉树,是树的基本操作之一,不算难题。最下面那句话实在有些木有节操啊,不知道是Google说给谁的。反正这道题确实难度不大,可以用递归和非递归两种方法来解。先来看递归的方法,写法非常简洁,五行代码搞定,交换当前左右节点,并直接调用递归即可,代码如下:

 

// Recursion
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (!root) return NULL;
        TreeNode *tmp = root->left;
        root->left = invertTree(root->right);
        root->right = invertTree(tmp);
        return root;
    }
};

 

非递归的方法也不复杂,跟二叉树的层序遍历一样,需要用queue来辅助,先把根节点排入队列中,然后从队中取出来,交换其左右节点,如果存在则分别将左右节点在排入队列中,以此类推直到队列中木有节点了停止循环,返回root即可。代码如下:

 

// Non-Recursion
class Solution {
public:
    TreeNode* invertTree(TreeNode* root) {
        if (!root) return NULL;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            TreeNode *node = q.front(); q.pop();
            TreeNode *tmp = node->left;
            node->left = node->right;
            node->right = tmp;
            if (node->left) q.push(node->left);
            if (node->right) q.push(node->right);
        }
        return root;
    }
};

 

LeetCode All in One 题目讲解汇总(持续更新中…)


转载请注明来源于 Grandyang 的博客 (grandyang.com),欢迎对文章中的引用来源进行考证,欢迎指出任何有错误或不够清晰的表达。可以在下面评论区评论,也可以邮件至 grandyang@qq.com

💰


微信打赏


Venmo 打赏

×

Help us with donation