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 does not need to go through the root.

For example:
Given the below binary tree,

       1
/ \
2 3

Return 6.

求最大路径。

就是记录两个结果。

/**
* 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) {
if( root == null)
return 0;
long result[] = helper(root); return (int)Math.max(result[0], result[1]);
} public long[] helper(TreeNode node){
long[] result = new long[2];
result[0] = Integer.MIN_VALUE;
result[1] = Integer.MIN_VALUE;
if( node == null )
return result;
result[0] = node.val;
result[1] = node.val;
if( node.left == null && node.right == null)
return result; long[] num1 = helper(node.left);
long[] num2 = helper(node.right); result[0] = Math.max(Math.max(num1[0],num2[0])+node.val,node.val);
result[1] = Math.max(Math.max(Math.max(Math.max(Math.max(num1[1],num2[1]),num1[0]+num2[0]+node.val),num1[0]+node.val),
num2[0]+node.val),node.val); return result;
}
}

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