leetcode-124. Binary Tree Maximum Path Sum

Posted 2022-12-11 千念飞羽

leetcode-124. Binary Tree Maximum Path Sum

题目：

Given a 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.

For example:
Given the below binary tree,

   1
  / \\
 2   3

Return 6.

这题比较有意思。这里计算的是所有可能路径的和。我一开始计算的是两个叶节点之间的最大和

这题显然是需要迭代的，但是每次迭代中却需要进行很多次比较，所以对于结果我使用了域。

针对一个节点需要比较这几个，
其左子节点对应的最大值，
其右子节点对应的最大值，
其从左子节点-根节点-右节点之和，
其从左子节点-根节点之和，
其根节点-右节点之和，
以及ret。

这几个值之间的最大值。

/**
 * Definition for a binary tree node.
 * public class TreeNode 
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x)  val = x; 
 * 
 */
public class Solution 
    private int ret = Integer.MIN_VALUE;
    public int maxPathSum(TreeNode root) 
        if(root==null) return Integer.MIN_VALUE;
        findmaxpath(root);
        return ret;
    

    private int findmaxpath(TreeNode node)
        if(node == null) return 0;
        int left = Math.max(0, findmaxpath(node.left));
        int right = Math.max(0, findmaxpath(node.right));
        ret = Math.max(ret, left + right + node.val);
        return node.val + Math.max(left,right);