[lintcode] Binary Tree Maximum Path Sum II

 

 

Given a binary tree, find the maximum path sum from root.

The path may end at any node in the tree and contain at least one node in it.

 

给一棵二叉树，找出从根节点出发的路径中，和最大的一条。

这条路径可以在任何二叉树中的节点结束，但是必须包含至少一个点（也就是根了）。

/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root the root of binary tree.
* @return an integer
*/
public int maxPathSum2(TreeNode root) {
if (root == null) {
return Integer.MIN_VALUE;
}

int left = maxPathSum2(root.left);
int right = maxPathSum2(root.right);

return root.val + Math.max(0, Math.max(left, right));
}
}