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.
Test Cases
Example 1:
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.
-10
/ \
9 20
/ \
15 7
Example 2:
Input: root = [1,2,3]
Output: 6
Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.
1
/ \
2 3
Solution
/**
* 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 maxVal[1] = {INT_MIN};
maxPathDown(root, maxVal);
return maxVal[0];
}
int maxPathDown(TreeNode* root, int* maxVal) {
if (root == nullptr) return 0;
int left = max(0, maxPathDown(root->left, maxVal));
int right = max(0, maxPathDown(root->right, maxVal));
maxVal[0] = max(maxVal[0], left+right+root->val);
return max(left, right) + root->val;
}
};/**
* 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 {
public int maxPathSum(TreeNode root) {
int[] maxVal = new int[]{Integer.MIN_VALUE};
maxPathDown(root, maxVal);
return maxVal[0];
}
private int maxPathDown(TreeNode root, int[] maxVal) {
if (root == null) return 0;
int left = Math.max(0, maxPathDown(root.left, maxVal));
int right = Math.max(0, maxPathDown(root.right, maxVal));
maxVal[0] = Math.max(maxVal[0], left+right+root.val);
return Math.max(left, right) + root.val;
}
}from typing import Optional, List
# 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:
max_value = [float('-inf')]
self.maxPathDown(root, max_value)
return int(max_value[0])
def maxPathDown(self, node: Optional[TreeNode], max_value: List[float]):
if not node:
return 0
left = max(0, self.maxPathDown(node.left, max_value))
right = max(0, self.maxPathDown(node.right, max_value))
max_value[0] = max(max_value[0], left + right + node.val)
return max(left, right) + node.val