Tree - 124. Binary Tree Maximum Path Sum

Published on with 0 views and 0 comments

124. Binary Tree Maximum Path Sum

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

思路:

题目意思是找出二叉树最大的一个路径和,这个路径不需要一定经过root节点,但是必须包含一个节点,题目还是比较阴险,因为如果都是负数,也得找出那个最大的,比如只有一个root节点,值为-1,也得返回-1,做法就是使用递归求解,如果root不为空,那么就找左右节点,看左右节点的从左走和从右走的值的最大值,同时记录下最大值,然后递归返回到父节点依次求解。注意递归返回的时候,不是返回最终结果的最大值,只是只经过一个子节点的那个路径的最大值。

代码:

go:

/**

 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
// import "math"

func maxPathSum(root *TreeNode) int {
    if root == nil {
        return math.MinInt32
    }

    var res = math.MinInt32
    dfs(root, &res)
    return res
}

func dfs(node *TreeNode, res *int) int {
    if node == nil {
        return 0
    }
 
    left := max(dfs(node.Left, res), 0) 
    right := max(dfs(node.Right, res), 0)
    
    // 记录最大路径和
    *res = max(*res, node.Val + left + right)
    
    return node.Val + max(left, right)
}


func max(i, j int) int {
    if i > j {
        return i
    }
    return j
}

标题:Tree - 124. Binary Tree Maximum Path Sum
作者:michael
地址:https://www.bitbo-liuyang.com/articles/2019/08/20/1566312646724.html