Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.
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/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isBalanced(TreeNode *root) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
int h = 0;
return isBalanced(root, h);
}
private:
bool isBalanced(TreeNode *root, int &h) {
if (root == NULL) {
h = 0;
return true;
}
if (root->left == NULL && root->right == NULL) {
h = 1;
return true;
}
int l = 0, r = 0;
bool ret = true;
if (!isBalanced(root->left, l) || !isBalanced(root->right, r)) ret = false;
h = max(l, r) + 1;
return ret && abs(l - r) <= 1;
}
};
本文介绍了一种算法来判断给定的二叉树是否为高度平衡树,即树的左右子树深度之差不超过1。通过递归方式实现判断逻辑,确保每个节点的子树深度信息被正确计算并比较。
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