LeetCode学习篇二十五——Balanced Binary Tree
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题目: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.
通过率:35.9% 难度:easy
这道题就是要判断所给的二叉树是不是平衡二叉树。平衡二叉树(Balanced Binary Tree)又被称为AVL树(有别于AVL算法),且具有以下性质:它是一 棵空树或它的左右两个子树的高度差的绝对值不超过1,并且左右两个子树都是一棵平衡二叉树。所以实现方法通过两层递归,一个递归是求树的高度,另一个递归是判断左右子树是否高度相差不超过1,代码实现如下:
/** * 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 height(TreeNode *root) { if(root == NULL) { return 0; } else return max(height(root->left),height(root->right))+1; } bool isBalanced(TreeNode* root) { if(root == NULL) { return true; } else if(height(root->left)-height(root->right) <= 1 && height(root->left)-height(root->right) >= -1) { return isBalanced(root->left) && isBalanced(root->right); } else return false; }};
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