原题
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.
判断一棵树是否是平衡二叉树，
注意的问题
1）：递归判断左右子树的平衡性(第12行)
2）：左右子树高度相差大于1，而不是等于2，这不是生成平衡树(第10行)
3）：平衡树是不是搜索树？这是一个问题。。。
1 class Solution 2 { 3 public: 4 bool isBalanced(TreeNode* root) 5 { 6 if (root == NULL) 7 return true; 8 int lh = height(root->left); 9 int rh = height(root->right); 10 if(lh - rh > 1 || rh - lh > 1) 11 return false; 12 return isBalanced(root->left)&&isBalanced(root->right); 13 } 14 //求节点高度 15 int height(TreeNode* &pos) 16 { 17 if(pos == NULL) 18 return -1; 19 //返回节点高度 20 return max(height(pos->left),height(pos->right)) + 1; 21 } 22 };
1 class Solution 2 { 3 public: 4 bool isBalanced(TreeNode* root) 5 { 6 if (root == NULL) 7 return true; 8 bool IsSearch = true; 9 int lh = height(root->left,IsSearch); 10 int rh = height(root->right,IsSearch); 11 if(lh - rh == 2 || rh - lh == 2 || !IsSearch) 12 return false; 13 return true; 14 } 15 //求节点高度 16 int height(TreeNode* &pos,bool &charge) 17 { 18 if(pos == NULL) 19 return -1; 20 if (pos->right != NULL) 21 { 22 if(pos->val >= pos->right->val) 23 { 24 charge = false; 25 return 0; 26 } 27 } 28 else if(pos->left != NULL) 29 { 30 if(pos->val <= pos->left->val) 31 { 32 charge = false; 33 return 0; 34 } 35 } 36 if(charge) 37 return max(height(pos->left,charge),height(pos->right,charge)) + 1; 38 else 39 return 0; 40 } 41 };