LeetCode 124. Binary Tree Maximum Path Sum（树中最长路径和，递归）

最新推荐文章于 2021-10-10 19:26:51 发布

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最新推荐文章于 2021-10-10 19:26:51 发布

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分类专栏：树 LeetCode

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本文探讨了在非空二叉树中寻找最大路径和的问题，定义路径为从任意节点开始到任意节点结束的序列，路径至少包含一个节点，不必通过根节点。通过递归方法，求解每个节点的最长路径和，最终返回整棵树的最大路径和。

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Given a non-empty binary tree, find the maximum path sum.

For this problem, a path is defined as any sequence of nodes from some starting node to any node in the tree along the parent-child connections. The path must contain at least one node and does not need to go through the root.

Example 1:

Input: [1,2,3]

       1
      / \
     2   3

Output: 6

Example 2:

Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

解法
递归思路：
求某个节点的最长路径和，只需要求左节点的最长路径和left，右节点的最长路径和right，以及当前节点的值val
maxpath = max(val, val+left, val+right, val+left+right)
递归函数返回的时候是一条链，因此返回max(val, val+left, val+right)

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def solve(self, root):
        if root:
            left = self.solve(root.left)
            right = self.solve(root.right)
            val = root.val
            ret = max(val, val+left, val+right) 
            self.ans = max(self.ans, ret, val+left+right)
            return ret
        return 0
        
    def maxPathSum(self, root):
        """
        :type root: TreeNode
        :rtype: int
        """
        self.ans = -(1<<30)
        self.solve(root)
        return self.ans