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.
//两次递归效率不高 合并看leetcode代码
/** * 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==NULL) return true; int leftDepth=getDepth(root->left); int rightDepth=getDepth(root->right); if(abs(leftDepth-rightDepth)>1) return false; else return isBalanced(root->left)&&isBalanced(root->right); }
//得到树的高度 int getDepth(TreeNode *node) { if(node==NULL) return 0; int leftDepth=getDepth(node->left); int rightDepth=getDepth(node->right); return 1+max(leftDepth,rightDepth); } };
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isBalanced(TreeNode* root) { 13 return balancedHeight(root)>=0; 14 15 } 16 17 //return the height of root if the tree is a balanced tree; 18 //otherwise,return -1 19 int balancedHeight(TreeNode *root) 20 { 21 if(root==NULL) return 0; 22 int left=balancedHeight(root->left); 23 int right=balancedHeight(root->right); 24 if(left<0||right<0||abs(left-right)>1) 25 return -1; 26 else 27 return max(left,right)+1; 28 } 29 };