Balanced Binary Tree——数是否是平衡,即任意节点左右字数高度差不超过1
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.
/**
* 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) {
if(root==NULL) return true;
int l=height(root->left);
int r=height(root->right);
if(l<||r<||l-r>||r-l>) return false;
return true;
}
int height(TreeNode *root){
if(root==NULL) return ;
int l=height(root->left);
int r=height(root->right);
if(l<||r<||l-r>||r-l>) return -;
return max(l,r)+;
}
};
其他做法:
转自:http://blog.csdn.net/feliciafay/article/details/18348065
// 68ms过大集合 简洁版
public:
int cntHeight(TreeNode *root) {
if(root == NULL) return ;
int l = cntHeight(root->left);
int r = cntHeight(root->right);
if(l < || r < || abs(l-r) > ) return -; //自定义 return -1,表示不平衡的情况
else return max(l, r) + ;
}
bool isBalanced(TreeNode *root) {
if(root == NULL) return true;
int l = cntHeight(root->left);
int r = cntHeight(root->right);
if(l < || r < || abs(l-r) > ) return false;
else return true;
}
// 68ms过大集合 更简洁版
public:
int cntHeight(TreeNode *root) {
if(root == NULL) return ;
int l = cntHeight(root->left);
int r = cntHeight(root->right);
if(l < || r < || abs(l-r) > ) return -; //自定义 return -1,表示不平衡的情况
else return max(l, r) + ;
}
bool isBalanced(TreeNode *root) {
if(root == NULL) return true;
int l = cntHeight(root->left);
int r = cntHeight(root->right);
if(l < || r < || abs(l-r) > ) return false;
else return true;
}
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