226. 翻转二叉树
已解答
简单
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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
思路:用递归来做
对每一层进行翻转,交换每一层的左右子树
C++代码实现:
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){return nullptr;}TreeNode *temp=root->left;root->left=root->right;root->right=temp;invertTree(root->left);invertTree(root->right);return root;}
};
JAVA代码实现:
Definition for a binary tree node.public class TreeNode {int val;TreeNode left;TreeNode right;TreeNode() {}TreeNode(int val) { this.val = val; }TreeNode(int val, TreeNode left, TreeNode right) {this.val = val;this.left = left;this.right = right;}}class Solution {public TreeNode invertTree(TreeNode root) {if(root==null){return null;}TreeNode temp=root.left;root.left=root.right;root.right=temp;invertTree(root.left);invertTree(root.right);return root;}
}