【LeetCode】226. Flip the binary tree

题目

给你一棵二叉树的根节点 root ,翻转这棵二叉树,并返回其根节点.

示例 1：

输入：root = [4,2,7,1,3,6,9]
输出：[4,7,2,9,6,3,1]

示例 2：

输入：root = [2,1,3]
输出：[2,3,1]

示例 3：

输入：root = []
输出：[]

提示：

树中节点数目范围在 [0, 100] 内
-100 <= Node.val <= 100

题解

使用递归
Swap from the bottom of the binary tree

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    TreeNode* invertTree(TreeNode* root) {
    
        if(root==nullptr)
            return nullptr;
        
        TreeNode* left = invertTree(root->left);
        TreeNode* right = invertTree(root->right);
        root->right = left;
        root->left = right;
        return root;
    }
};

Swap from top to bottom

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    void fun(TreeNode* root)
    {
    
        TreeNode* tmp = root->left;
        root->left = root->right;
        root->right = tmp;
        if(root->left)
            fun(root->left);
        if(root->right)
            fun(root->right);
    }
    TreeNode* invertTree(TreeNode* root) {
    
        if(root==nullptr)
            return nullptr;
        
        fun(root);
        return root;
    }
};