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.
root of a binary tree, return the maximum path sum of any path.
Input: root = [1,2,3] Output: 6 Explanation: The optimal path is 2 -> 1 -> 3 with a path sum of 2 + 1 + 3 = 6.
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.
- The number of nodes in the tree is in the range
[1, 3 * 104].
-1000 <= Node.val <= 1000
Use a helper function to find the max path value from a node. Traverse each node to globally find the maximum path sum of the whole tree.
# 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: TreeNode) -> int: self.max_sum = -sys.maxsize self.helper(root) return self.max_sum def helper(self, root): if root == None: return 0 left = max(self.helper(root.left), 0) right = max(self.helper(root.right), 0) self.max_sum = max(self.max_sum, left + right + root.val) return max(left, right) + root.val
- Time Complexity: O(N).
- Space Complexity: O(N).