# brute force and bottom up

• ``````/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
//Brute force
//O(N ^ 2)
//O(N)
class Solution {
public:
//divide & conquer
bool isBalanced(TreeNode* root) {
if (root == NULL) return true;
return isBalanced(root->left) &&
isBalanced(root->right) &&
abs(height(root->left) - height(root->right)) <= 1;
}

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

O(N)
O(N)
/**
* 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:
//brute force: repetition calculation of depth
// how to avoid it

//1 pass depth bottom-up
//2 use -1 to indicate unbalanced
bool isBalanced(TreeNode* root) {
return balance(root) != -1;
}

int balance(TreeNode* root) {
if (root == NULL) return 0;
int left = balance(root->left);
int right = balance(root->right);
if (left == -1 ||
right == -1 ||
abs(left - right) > 1)
return -1;

return max(left, right) + 1;
}
};
``````

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