Solution 1DFS Recursive
# 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:
self.maxVal = float('-inf')
self.nodeMax(root)
return self.maxVal
def nodeMax(self, node) -> int:
if node == None:
return float('-inf')
left = self.nodeMax(node.left)
right = self.nodeMax(node.right)
nodeMax = max(node.val + left, node.val + right, node.val)
self.maxVal = max(self.maxVal, nodeMax, node.val + left + right)
return nodeMax/**
* 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;
* }
* }
*/
int max = -10000;
public int maxPathSum(TreeNode root) {
maxWithChild(root);
return max;
}
public int maxWithChild(TreeNode root) {
if (root == null) return -10000;
int left = maxWithChild(root.left);
int right = maxWithChild(root.right);
int nodeMax = Math.max(root.val, root.val + Math.max(left, right));
max = Math.max(max, Math.max(nodeMax, root.val + left + right));
return nodeMax;
}