124. Binary Tree Maximum Path Sum

Hard (Khó) Python 🔗 Xem trên LeetCode

📋 Đề Bài

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

🧠 Thuật Toán & Kỹ Thuật

DFS (Tìm kiếm theo chiều sâu)Tree Traversal (Duyệt cây)
⏱️ Thời gian O(V+E)
💾 Không gian O(V)

💻 Lời Giải

Python 0124-binary-tree-maximum-path-sum.py
# 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.maxValue = -inf
        
        def dfs(root):
            if not root:
                return 0
            l = dfs(root.left)
            r = dfs(root.right)
            self.maxValue = max(self.maxValue, root.val + l + r)
            return max(root.val + max(l, r), 0)
        
        dfs(root)
        return self.maxValue