Binary Tree Maximum Path Sum Problem
Description
LeetCode Problem 124.
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:
1
2
3
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:
1
2
3
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 * 10^4].
- -1000 <= Node.val <= 1000
Sample C++ Code
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
class Solution {
public:
int dfsMaxSum(TreeNode* root, int &max_sum) {
if (root == NULL)
return 0;
int left_sum = dfsMaxSum(root->left, max_sum);
int right_sum = dfsMaxSum(root->right, max_sum);
int temp = max(max(left_sum, right_sum) + root->val, root->val);
max_sum = max(max_sum, max(temp, left_sum + right_sum + root->val));
return temp;
}
int maxPathSum(TreeNode* root) {
if (root == NULL)
return 0;
int max_sum = root->val;
int val = dfsMaxSum(root, max_sum);
return max_sum;
}
};