LEETCODE-Invert Binary Tree

最新推荐文章于 2022-02-12 15:15:58 发布

prettysky123

最新推荐文章于 2022-02-12 15:15:58 发布

阅读量297

点赞数

CC 4.0 BY-SA版权

分类专栏： leetcode 文章标签：二叉树 inverttree

本文链接：https://blog.youkuaiyun.com/prettysky123/article/details/49664751

leetcode 专栏收录该内容

137 篇文章

订阅专栏

本文详细介绍了反转二叉树的两种实现方式：递归实现和非递归实现，通过具体代码实例帮助理解算法原理。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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.
方法一：递归实现

/**
 * 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)
         return NULL;
       TreeNode* r;
       r = root->left;
       root->left = invertTree(root->right);
       root->right = invertTree(r);
    }
};

方法二：利用队列（queue）非递归实现

/**
 * 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* Change(TreeNode* needl){
        TreeNode* temp;
        temp = needl->left;
        needl->left = needl->right;
        needl->right = temp;
        return needl;
    }
    TreeNode* invertTree(TreeNode* root) {
       queue<TreeNode*> BiQueue;
       if(!root)
         return NULL;
        BiQueue.push(root);
        TreeNode* needl;
        while(!BiQueue.empty()){
            needl = BiQueue.front();
            BiQueue.pop();
            Change(needl);
            if(needl->left)
             BiQueue.push(needl->left);
            if(needl->right)
             BiQueue.push(needl->right);
        }
        return root;
    }
};