LeetCode226翻转二叉树

Posted 2020-12-18 臭咸鱼

题目链接

https://leetcode-cn.com/problems/invert-binary-tree/

题解一

• 递归解法
• 我写的，不够简洁

// Problem: LeetCode 226
// URL: https://leetcode-cn.com/problems/invert-binary-tree/
// Tags: Tree Recursion
// Difficulty: Easy

#include <iostream>
using namespace std;

struct TreeNode{
    TreeNode* left;
    TreeNode* right;
    int val;
    TreeNode(int x):val(x),left(nullptr),right(nullptr){}
};

class Solution{
public:
    TreeNode* invertTree(TreeNode* root) {
        // 递归出口，空结点直接返回
        if(root==nullptr)
            return nullptr;
        // 该结点交换左右子树
        TreeNode *temp = root->left;
        root->left = root->right;
        root->right = temp;
        // 递归翻转左右子树
        invertTree(root->left);
        invertTree(root->right);
        return root;
    }
};

题解二

• 比我写的简洁一些，交换了最后两步的顺序，特别是利用了二叉树翻转后根结点不变的特点

// Problem: LeetCode 226
// URL: https://leetcode-cn.com/problems/invert-binary-tree/
// Tags: Tree Recursion
// Difficulty: Easy

#include <iostream>
using namespace std;

struct TreeNode{
    TreeNode* left;
    TreeNode* right;
    int val;
    TreeNode(int x):val(x),left(nullptr),right(nullptr){}
};

class Solution{
public:
    TreeNode* invertTree(TreeNode* root) {
        // 递归出口，空结点直接返回
        if(root==nullptr)
            return nullptr;
        // 递归翻转左右子树
        TreeNode* left = invertTree(root->left);
        TreeNode* right = invertTree(root->right);
        // 该结点交换左右子树
        root->left = right;
        root->right = left;
        return root;
    }
};

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作者：@臭咸鱼

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