> For the complete documentation index, see [llms.txt](https://coding-9.gitbook.io/untitled/llms.txt). Markdown versions of documentation pages are available by appending `.md` to page URLs; this page is available as [Markdown](https://coding-9.gitbook.io/untitled/trees/binary-tree/hard/124.-binary-tree-maximum-path-sum.md).

# 124. Binary Tree Maximum Path Sum

## Problem Statement

<br>

A **path** in a binary tree is a sequence of nodes where each pair of adjacent nodes in the sequence has an edge connecting them. A node can only appear in the sequence **at most once**. Note that the path does not need to pass through the root.

The **path sum** of a path is the sum of the node's values in the path.

Given the `root` of a binary tree, return *the maximum **path sum** of any **non-empty** path*.

&#x20;

**Example 1:**

![](https://assets.leetcode.com/uploads/2020/10/13/exx1.jpg)

<pre><code><strong>Input: root = [1,2,3]
</strong><strong>Output: 6
</strong><strong>Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.
</strong></code></pre>

**Example 2:**

![](https://assets.leetcode.com/uploads/2020/10/13/exx2.jpg)

<pre><code><strong>Input: root = [-10,9,20,null,null,15,7]
</strong><strong>Output: 42
</strong><strong>Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.
</strong></code></pre>

&#x20;

**Constraints:**

* The number of nodes in the tree is in the range `[1, 3 * 104]`.
* `-1000 <= Node.val <= 1000`

## Intuition

```
Approach:

At each step: Take either left or right
or left + right + root(current)


And in the parent we return, Max(l,r) + root
```

### Links

<https://leetcode.com/problems/binary-tree-maximum-path-sum/description/>

### Video Links

<https://leetcode.com/problems/binary-tree-maximum-path-sum/solutions/603072/c-solution-o-n-with-detailed-explanation/>

### Approach 1:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
class Solution {
public:
    int max_sum=INT_MIN;
    int max_gain(TreeNode* root)
    {
        if(!root)
            return 0;
        int l=max(max_gain(root->left),0);
        int r=max(max_gain(root->right),0);

        int new_price=root->val+l+r;
        max_sum=max(max_sum,new_price);

        return root->val+max(l,r);
    }
    int maxPathSum(TreeNode* root) {
        max_gain(root);
        
        return max_sum;
    }
};
```

{% endcode %}

### Approach 2:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 3:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Approach 4:

```
```

{% code title="C++" lineNumbers="true" %}

```cpp
```

{% endcode %}

### Similar Problems

###