【二叉树经典问题】110. Balanced Binary Tree
2017-08-25 23:06
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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.
解答:
写递归就是跟着感觉走,美滋滋:
/**
* 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:
bool isBalanced(TreeNode* root) {
if(!root) return true;
return (abs(get_height(root->left)-get_height(root->right))<=1)&&(isBalanced(root->left))&&(isBalanced(root->right));
}
int get_height(TreeNode* root){
if(!root)return 0;
int l=get_height(root->left);
int r=get_height(root->right);
return l>r?(l+1):(r+1);
//return root==NULL?0:1+max(get_height(root->left),get_height(root->right));
}
};倒是求树高的代码要注意
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.
解答:
写递归就是跟着感觉走,美滋滋:
/**
* 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:
bool isBalanced(TreeNode* root) {
if(!root) return true;
return (abs(get_height(root->left)-get_height(root->right))<=1)&&(isBalanced(root->left))&&(isBalanced(root->right));
}
int get_height(TreeNode* root){
if(!root)return 0;
int l=get_height(root->left);
int r=get_height(root->right);
return l>r?(l+1):(r+1);
//return root==NULL?0:1+max(get_height(root->left),get_height(root->right));
}
};倒是求树高的代码要注意
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