【笔记】力扣 110. 平衡二叉树
简单
给定一个二叉树,判断它是否是 平衡二叉树
示例 1:

输入:root = [3,9,20,null,null,15,7]
输出:true
示例 2:

输入:root = [1,2,2,3,3,null,null,4,4]
输出:false
示例 3:
输入:root = []
输出:true
提示:
- 树中的节点数在范围
[0, 5000]内 -104 <= Node.val <= 104
题解
/**
* 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:
bool dfs(TreeNode* root) {
if (root==nullptr) return true;
if (root->left==nullptr&&root->right==nullptr) return true;
return dfs(root->left)&&dfs(root->right)&&(abs(dfs2(root->left)-dfs2(root->right))<=1);
}
int dfs2(TreeNode* root) {
if (root==nullptr) return 0;
return max(dfs2(root->left),dfs2(root->right))+1;
}
bool isBalanced(TreeNode* root) {
return dfs(root);
}
};
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