# 110. Balanced Binary Tree

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.

Example 1:

Given the following tree [3,9,20,null,null,15,7]:

    3
   / \
  9  20
    /  \
   15   7

Return true.

Example 2:

Given the following tree [1,2,2,3,3,null,null,4,4]:

       1
      / \
     2   2
    / \
   3   3
  / \
 4   4

Return false.

# Solution

Approach 1: Recursion. To avoid the tricky case that a result is determined by 2 separate things (left and right's balance and their depth), we can use -1 to represent the "depth" when the node is not balanced.

# Code (Python)

Approach 1:

    def isBalanced(self, root):
        return self._helper(root) != -1
    
    def _helper(self, node):
        if not node:
            return 0
        left, right = self._helper(node.left), self._helper(node.right)
        if left == -1 or right == -1 or abs(left - right) > 1:
            return -1
        return max(left, right) + 1

# Code (C++)

Approach 1:

/**
 * 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) {
        int height = 0;
        return isBalanced(root, height);
    }
private:
    bool isBalanced(TreeNode *root, int& height)
    {
        if (root == NULL) return true;
        int leftHeight = 0;
        int rightHeight = 0;
        bool leftIsBalanced = isBalanced(root->left, leftHeight);
        bool rightIsBalanced = isBalanced(root->right, rightHeight);
        height = std::max(leftHeight, rightHeight) + 1;
        if (leftIsBalanced && rightIsBalanced && abs(leftHeight - rightHeight) <= 1)
            return true;
        return false;
    }
};