# 124. Binary Tree Maximum Path Sum
## Problem Statement
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*.
**Example 1:**
Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.
**Example 2:**
Input: root = [-10,9,20,null,null,15,7]
Output: 42
Explanation: The optimal path is 15 -> 20 -> 7 with a path sum of 15 + 20 + 7 = 42.
**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
### Video Links
### 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
###
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