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
Solution
C++
/**
* 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 dfs(int& res, TreeNode* node) {
if (node == nullptr) return 0;
int val = node->val;
int left = max(0, dfs(res, node->left)), right = max(0, dfs(res, node->right));
res = max(res, val + left + right);
return val + max(0, max(left, right));
}
int maxPathSum(TreeNode* root) {
int res = INT_MIN;
dfs(res, root);
return res;
}
};Python3
# 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:
res = -inf
def go(node):
nonlocal res
if not node: return 0
curr = node.val
left, right = max(0, go(node.left)), max(0, go(node.right))
res = max(res, node.val + left + right)
return node.val + max(0, left, right)
go(root)
return res