Leetcode: Binary Tree Maximum Path Sum

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.

难度：75.

参见了他人的思路，这道题是求树的路径和的题目，不过和平常不同的是这里的路径不仅可以从根到某一个结点，而且路径可以从左子树某一个结点，然后到达右子树的结点，就像题目中所说的可以起始和终结于任何结点。函数的返回值定义为以自己为根的一条从根到叶子结点的最长路径，这个返回值是为了提供给它的父结点计算自身的最长路径用。这样一来，一个结点自身的最长路径就是它的左子树返回值（如果大于0的话），加上右子树的返回值（如果大于0的话），再加上自己的值。在过程中求得当前最长路径时比较一下是不是目前最长的，如果是则更新。算法的本质还是一次树的遍历，所以复杂度是O(n)。而空间上仍然是栈大小O(logn)。注意这里path的存值方式是个问题，如果用一个integer变量代入recursion的话，函数无法对实参造成改变，所以用了一个对象ArrayList<Integer> res的第一个元素来存最大的path值

 1 /**
 2  * Definition for binary tree
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 public class Solution {
11     public int maxPathSum(TreeNode root) {
12         if (root == null) return 0;
13         ArrayList<Integer> res = new ArrayList<Integer>();
14         res.add(Integer.MIN_VALUE);
15         FindMaxSum(root, res);
16         return res.get(0);
17     }
18
19     public int FindMaxSum(TreeNode root, ArrayList<Integer> res) {
20         if (root == null) {
21             return 0;
22         }
23         int leftsum = FindMaxSum(root.left, res);
24         int rightsum = FindMaxSum(root.right, res);
25         int maxsum = root.val + (leftsum>0? leftsum : 0) + (rightsum>0? rightsum : 0);
26         if (maxsum > res.get(0)) res.set(0, maxsum);
27         return root.val + Math.max(leftsum, Math.max(rightsum, 0));
28     }
29 }

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