# 124. Binary Tree Maximum Path Sum

## Description

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

## Solutions

Solution 1: Recursion

When thinking about the classic routine of recursion problems in binary trees, we consider:

1. Termination condition (when to terminate recursion)
2. Recursively process the left and right subtrees
3. Merge the calculation results of the left and right subtrees

For this problem, we design a function $dfs(root)$, which returns the maximum path sum of the binary tree with $root$ as the root node.

The execution logic of the function $dfs(root)$ is as follows:

If $root$ does not exist, then $dfs(root)$ returns $0$;

Otherwise, we recursively calculate the maximum path sum of the left and right subtrees of $root$, denoted as $left$ and $right$. If $left$ is less than $0$, then we set it to $0$, similarly, if $right$ is less than $0$, then we set it to $0$.

Then, we update the answer with $root.val + left + right$. Finally, the function returns $root.val + \max(left, right)$.

In the main function, we call $dfs(root)$ to get the maximum path sum of each node, and the maximum value among them is the answer.

The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.

• /**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode() {}
*     TreeNode(int val) { this.val = val; }
*     TreeNode(int val, TreeNode left, TreeNode right) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
class Solution {
private int ans = -1001;

public int maxPathSum(TreeNode root) {
dfs(root);
return ans;
}

private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int left = Math.max(0, dfs(root.left));
int right = Math.max(0, dfs(root.right));
ans = Math.max(ans, root.val + left + right);
return root.val + Math.max(left, right);
}
}

• /**
* 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 {
public:
int maxPathSum(TreeNode* root) {
int ans = -1001;
function<int(TreeNode*)> dfs = [&](TreeNode* root) {
if (!root) {
return 0;
}
int left = max(0, dfs(root->left));
int right = max(0, dfs(root->right));
ans = max(ans, left + right + root->val);
return root->val + max(left, right);
};
dfs(root);
return ans;
}
};

• # Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
def maxPathSum(self, root: Optional[TreeNode]) -> int:
def dfs(root: Optional[TreeNode]) -> int:
if root is None:
return 0
left = max(0, dfs(root.left))
right = max(0, dfs(root.right))
nonlocal ans
ans = max(ans, root.val + left + right)
return root.val + max(left, right)

ans = -inf
dfs(root)
return ans


• /**
* Definition for a binary tree node.
* type TreeNode struct {
*     Val int
*     Left *TreeNode
*     Right *TreeNode
* }
*/
func maxPathSum(root *TreeNode) int {
ans := -1001
var dfs func(*TreeNode) int
dfs = func(root *TreeNode) int {
if root == nil {
return 0
}
left := max(0, dfs(root.Left))
right := max(0, dfs(root.Right))
ans = max(ans, left+right+root.Val)
return max(left, right) + root.Val
}
dfs(root)
return ans
}

• /**
* Definition for a binary tree node.
* class TreeNode {
*     val: number
*     left: TreeNode | null
*     right: TreeNode | null
*     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
*         this.val = (val===undefined ? 0 : val)
*         this.left = (left===undefined ? null : left)
*         this.right = (right===undefined ? null : right)
*     }
* }
*/

function maxPathSum(root: TreeNode | null): number {
let ans = -1001;
const dfs = (root: TreeNode | null): number => {
if (!root) {
return 0;
}
const left = Math.max(0, dfs(root.left));
const right = Math.max(0, dfs(root.right));
ans = Math.max(ans, left + right + root.val);
return Math.max(left, right) + root.val;
};
dfs(root);
return ans;
}


• /**
* Definition for a binary tree node.
* function TreeNode(val, left, right) {
*     this.val = (val===undefined ? 0 : val)
*     this.left = (left===undefined ? null : left)
*     this.right = (right===undefined ? null : right)
* }
*/
/**
* @param {TreeNode} root
* @return {number}
*/
var maxPathSum = function (root) {
let ans = -1001;
const dfs = root => {
if (!root) {
return 0;
}
const left = Math.max(0, dfs(root.left));
const right = Math.max(0, dfs(root.right));
ans = Math.max(ans, left + right + root.val);
return Math.max(left, right) + root.val;
};
dfs(root);
return ans;
};


• /**
* Definition for a binary tree node.
* public class TreeNode {
*     public int val;
*     public TreeNode left;
*     public TreeNode right;
*     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
*         this.val = val;
*         this.left = left;
*         this.right = right;
*     }
* }
*/
public class Solution {
private int ans = -1001;

public int MaxPathSum(TreeNode root) {
dfs(root);
return ans;
}

private int dfs(TreeNode root) {
if (root == null) {
return 0;
}
int left = Math.Max(0, dfs(root.left));
int right = Math.Max(0, dfs(root.right));
ans = Math.Max(ans, left + right + root.val);
return root.val + Math.Max(left, right);
}
}

• // Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, res: &mut i32) -> i32 {
if root.is_none() {
return 0;
}
let node = root.as_ref().unwrap().borrow();
let left = (0).max(Self::dfs(&node.left, res));
let right = (0).max(Self::dfs(&node.right, res));
*res = (node.val + left + right).max(*res);
node.val + left.max(right)
}

pub fn max_path_sum(root: Option<Rc<RefCell<TreeNode>>>) -> i32 {
let mut res = -1000;
Self::dfs(&root, &mut res);
res
}
}