leetcode-124-Binary-Tree-Maximum-Path-Sum

发表于   |  

描述

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:

1
2
3
4
5
6
7
Input: [1,2,3]

       1
      / \
     2   3

Output: 6

Example 2:

1
2
3
4
5
6
7
8
9
Input: [-10,9,20,null,null,15,7]

   -10
   / \
  9  20
    /  \
   15   7

Output: 42

分析

二叉树里路径可以从任意节点开始到任意节点结束，二叉树可以用 DFS 来遍历，先算出左右子树的结果 left, right, 如果大于0，则对结果有利，加上 left 或 right 后与之前的 max 进行比较，取较大值。

解决方案1（Python）

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
# 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 __init__(self):
        self.maxSum = float("-inf")
        
    def maxPathSum(self, root: Optional[TreeNode]) -> int:
        def dfs(node):
            if not node:
                return 0
            left = max(dfs(node.left), 0)
            right = max(dfs(node.right), 0)
            
            self.maxSum = max(self.maxSum, node.val + left + right)
            return node.val + max(left, right)
        dfs(root)
        return self.maxSum

解决方案1(Java)

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    int max = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) {
        dfs(root);
        return max;
    }
    
    private int dfs(TreeNode root) {
        if (root == null) {
            return 0;
        }
        int left = Math.max(dfs(root.left), 0);
        int right = Math.max(dfs(root.right), 0);
        max = Math.max(max, root.val + left + right);
        return root.val + Math.max(left, right);
    }
}

相关问题

题目来源