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.
这个题目的意思是,如果一个二叉树树的任何节点的左右子树的深度之差不大于1,就是一个高度平衡的二叉树,求你判断一棵树是否是高度平衡的二叉树
我的思路是,如果遍历到某个节点时,它的左右子树深度之差大于1,则返回-1,否则返回以这个节点为跟时树的高度,这样遍历完全后,即可判断是否是平衡的二叉树
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int depthFirstSearch(TreeNode *root)
{
if (root == NULL)return 0;
int leftdep = depthFirstSearch(root->left);
int rightdep=depthFirstSearch(root->right);
if (abs(leftdep - rightdep) > 1|| leftdep == -1 || rightdep == -1)return -1;
return leftdep > rightdep ? leftdep + 1 : rightdep + 1;
}
bool isBalanced(TreeNode* root) {
if (depthFirstSearch(root) == -1)return false;
else return true;
}
};