[leedcode 124] 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.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    int max;
/*    getMax函数的意思是经过node节点的最大和。求最大和，有以下四种情况：
    1. Node only （因为本题中的节点可能是负值！）
    2. L-sub + Node
    3. R-sub + Node
    4. L-sub + Node + R-sub
    求最大的结果，需要一个全局变量max，不断更新此变量
    */
    public int maxPathSum(TreeNode root) {
        max=Integer.MIN_VALUE;
        getMax(root);
        return max;
        
    }
    public int getMax(TreeNode node){
        if(node==null) return 0;
        int left=getMax(node.left);
        int right=getMax(node.right);
        int resMax=node.val;
        if(left>0) resMax+=left;
        if(right>0) resMax+=right;
        max=Math.max(max,resMax);
        return Math.max(node.val,Math.max(left+node.val,right+node.val));
    }
}