[Leetcode] [Java] Binary Tree Maximum Path Sum

Title:

Given a binary tree, find the maximum path sum.

The path may start and end at any node in the tree.

For example:
Given the below binary tree,

       1      /      2   3

Return 6 .

Test Instructions:

Given a binary tree, find the maximum path and.

The starting and ending positions of a path can be arbitrary nodes in the tree.

Example

Given a binary tree such as the following

       1      /      2   3

return 6 .

Algorithm Analysis:

1) recursively solve this problem

2) Get largest left sum and right sum

3) Compare to the stored maximum

* Here's a place to be clear, the shortest path result has four possible, current plus left, current plus right. Currently, the return value of the current plus left and right subtree

* and return value, there are three possible, current plus left, current plus right. Present themselves.

* Here return is not the current subtree, that is, the left and right add up ... Because it is only a road, it does not meet the requirements of the topic.

* Here, the return value of the function is defined as the longest path from the root to the child node with its own root

AC Code:

<span style= "Font-family:microsoft yahei;font-size:12px;" >/** * Definition for a binary tree node.  * public class TreeNode {*     int val, *     TreeNode left, *     TreeNode right; *     TreeNode (int x) {val = x;} *} */public class Solution {public    int maxpathsum (TreeNode root)     {    int max[] = new int[1];     Max[0] = Integer.min_value;    Calculatesum (Root, max);    return max[0];    }         public int calculatesum (TreeNode root, int[] max)     {    if (root = null)    return 0;         int left = Calculatesum (Root.left, max);    int right = Calculatesum (Root.right, max);         int current = Math.max (Root.val, Math.max (Root.val + left, Root.val + right));         Max[0] = Math.max (max[0], Math.max (current, left + Root.val + right));         return current;    }} </span>

[Leetcode] [Java] Binary Tree Maximum Path Sum