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
Approach - DFS
- There are three possible answer after we compute left and right, one is left + current + right other is left + current and right + current
Time: O(n) Space: O(h)
class Solution {
int max = Integer.MIN_VALUE;
public int maxPathSum(TreeNode root) {
calculate(root);
return max;
}
public int calculate(TreeNode root) {
if (root == null)
return 0;
int left = Math.max(calculate(root.left),0);
int right = Math.max(calculate(root.right),0);
max = Math.max(max, root.val + left + right);
return root.val + Math.max(left,right);
}
}