/*
题目: 树中最大路径和
难度: Easy
链接: http://www.itint5.com/oj/#13
问题:
给定一棵树的根结点，树中每个结点都包含一个整数值val。
我们知道树中任意2个结点之间都存在唯一的一条路径，路径值为路径上所有结点值之和。
请计算最大的路径值（允许路径为空）。
样例：
-10
/ | \
2 3 4
/ \
5 -1
/
6
/
-1
最大的路径值为13，相应的路径为5到6之间的路径。
扩展：此题算法也可用来解决另一个非常常见的面试题“树的直径”（求树中任意两结点路径的长度的最大值）。
可以认为树中每个结点的val值为1，那么求最长路径相当于求路径值最大的路径。
Solution: LeetCode中Binary Tree Maximum Path Sum的扩展，这里是树，而非二叉树。
code source from blog:
https://github.com/AnnieKim/ITint5/blob/master/013_%E6%A0%91%E4%B8%AD%E6%9C%80%E5%A4%A7%E8%B7%AF%E5%BE%84%E5%92%8C.cpp
*/
public static int maxTreePathSumRe(TreeNode root, ref int res)
{
if (root == null) return res;
int N = (root.children==null)?0:root.children.Length;
// julia added comment: need to find maximum path through the root node,
// so, multiple children, only keep first two most biggest pathes.
// For the test case, root.val = -10, there are 3 children, each has maximum path length through the root node: 2, 8, 4
// so, the consideration is first 2 largest path length, 8 and 4.
// -10 + 8 +4
// yeah, through the debug, she knows the idea.
int first = 0, second = 0;
for (int i = 0; i < N; ++i)
{
TreeNode[] a = root.children;
TreeNode b = (TreeNode)a[i];
int sum = maxTreePathSumRe(b, ref res);
if (sum > first) {
second = first;
first = sum;
} else if (sum > second) {
second = sum;
}
}
int maximum = root.val;
maximum = max(maximum, root.val + first);
res = max(res, maximum);
res = max(res, root.val + first + second);
return maximum;
}
public static int maxTreePathSum(TreeNode root)
{
int res = 0;
maxTreePathSumRe(root, ref res);
return res;
}
static void Main(string[] args)
{
/*
* Test case:
*
-10
/ | \
2 3 4
/ \
5 -1
/
6
/
-1
* Maximum path sum: 5-> 3-> -1-> 6 , result is 13
*/
TreeNode n1 = new TreeNode(-10);
n1.children = new TreeNode[3];
n1.children[0] = new TreeNode(2);
n1.children[1] = new TreeNode(3);
n1.children[2] = new TreeNode(4);
TreeNode l2 = n1.children[1];
l2.children = new TreeNode[2];
l2.children[0] = new TreeNode(5);
l2.children[1] = new TreeNode(-1);
TreeNode l3 = l2.children[1];
l3.children = new TreeNode[1];
l3.children[0] = new TreeNode(6);
l3.children[0].children = new TreeNode[1];
l3.children[0].children[0] = new TreeNode(-1);
int result = maxTreePathSum(n1);
}
