**source:**https://leetcode.com/problems/balanced-binary-tree**C/C++**

**Solution to LeetCode**problem

**110**.

**Balanced Binary Tree**.

## Problem

Given a binary tree, determine if it is **height-balanced**.

A **height-balanced** binary tree is a binary tree in which the depth of the two subtrees of every node never differs by more than one.

## Examples

**Example 1:**

Input:root = [3,9,20,null,null,15,7]

Output:true

**Example 2:**

Input:root = [1,2,2,3,3,null,null,4,4]

Output:false

**Example 3:**

Input:root = []

Output:true

## Constraints

- The number of nodes in the tree is in the range
`[0, 5000]`

. `-10`

^{4}<= Node.val <= 10^{4}

## Solution

We will use recursion to perform a DFS (Depth First Search) in order to determine the heigh of each subtree.

- We call the recursive method with the root node.
- If the current node doesn’t exist,
`return 0`

. - Explore the left and right nodes.
- If the absolute difference of nodes’ depth is higher than 1, we return -1.
- If the the depth of the left or right node is -1, we return -1 for the current node.
- Otherwise, we return the max hight of the nodes (left vs right) plus 1.

- If the current node doesn’t exist,
- If the final
`result == -1`

then the tree is unbalanced.

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/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
private:
int exploreNode(TreeNode* node) {
if (node == nullptr) return 0;
int lH = exploreNode(node->left);
int rH = exploreNode(node->right);
if (lH == -1 || rH == -1 || abs(lH - rH) > 1)
return -1;
return max(lH, rH) + 1;
}
public:
bool isBalanced(TreeNode* root) {
int r = exploreNode(root);
return r != -1;
}
};