leetcode每日一题-110:平衡二叉树
链接
平衡二叉树
题目
分析
遍历树,然后每次判断树的左右两个子树的差值即可。
代码
C++
/**
* 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 flag = 1;
bool isBalanced(TreeNode* root) {
dfs(root);
return flag;
}
int dfs(TreeNode* root) {
if (root == nullptr) {
return 0;
}
int left = dfs(root->left), right = dfs(root->right);
if (abs(left - right) > 1) {
flag = 0;
}
return 1 + max(left, right);
}
};
Java
class Solution {
public boolean isBalanced(TreeNode root) {
if (root == null) {
return true;
} else {
return Math.abs(height(root.left) - height(root.right)) <= 1 && isBalanced(root.left) && isBalanced(root.right);
}
}
public int height(TreeNode root) {
if (root == null) {
return 0;
} else {
return Math.max(height(root.left), height(root.right)) + 1;
}
}
}
作者:LeetCode-Solution