LeetCode Balanced Binary Tree

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isBalanced(TreeNode *root) {
int h;
return dfs(root, h);
}

bool dfs(TreeNode* root, int& height) {
height = 0;
if (root == NULL) {
return true;
}
int lh, rh;
if (!dfs(root->left, lh)) return false;
if (!dfs(root->right, rh)) return false;
lh++, rh++;
int diff = lh - rh;
if (diff < -1 || diff > 1) return false;
if (diff < 0) {
height =  rh;
} else {
height =  lh;
}
return true;
}
};

第二轮：

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.

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
bool valid;
public:
bool isBalanced(TreeNode *root) {
valid = true;
height(root);
return valid;
}

int height(TreeNode* root) {
if (!valid || root == NULL) {
return 0;
}
int hl = height(root->left);
int hr = height(root->right);
int diff = max(hl, hr) - min(hl, hr);
if (diff > 1) {
valid = false;
}
return max(hl, hr) + 1;
}
};

再简化一下：

/**
* 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:
bool isBalanced(TreeNode* root) {
return height(root) >= 0;
}

int height(TreeNode* root) {
if (root == NULL) {
return 0;
}
int lh = height(root->left);
int rh = height(root->right);
if (lh < 0 || rh < 0 || abs(lh - rh) > 1) {
return -1;
}
return 1 + max(lh, rh);
}
};