Hard
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
- compare max at each node
- left call, right call
- compare max with node's val+left val, node's val+right val, node's val+left val+right val
- return max of node's val, node's val+left val, node's val+right 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 {
int max;
public int maxPathSum(TreeNode root) {
max = -1001;
helper(root);
return max;
}
private int helper(TreeNode root) {
if(root == null)
return 0;
int val = root.val;
max = Math.max(max, val);
int lVal = helper(root.left);
int rVal = helper(root.right);
max = Math.max(max, Math.max(rVal+val, Math.max(lVal+val, lVal+rVal+val)));
return Math.max(rVal+val, Math.max(lVal+val, val));
}
}- Time Complexity: O(n)
- Space Complexity: O(logn), recursive call stacks