LeetCode 110. Balanced Binary Tree
2016-09-05 14:55
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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.
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
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; } };
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