leetcode-124:Binary Tree Maximum Path Sum(Java)

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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,

要求：计算二叉树中的路径之和最大值，起止节点为任意

解法：动态规划，对每一个节点，以该节点为根节点的最大值设为value，其左子树的路径最大值为lmax，右子树的路径最大值为rmax，(!!! 注意：这里的lmax和rmax指的是从左/右子节点出发的某一条单向路径，例如对于节点4，lmax=2+3 rmax=6+7+8,而并不是lmax=1+2+3 rmax=5+6+7+8)那么有：

value = value + (lmax>0?lmax:0) + (rmax>0?rmax:0) ;

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */import java.math.*;public class Solution {    int maxSum = Integer.MIN_VALUE;        public int maxPathSum(TreeNode root) {        if(root == null)            return 0;                    getMaxSumWithCurNode(root);        return maxSum;    }            int getMaxSumWithCurNode(TreeNode curNode){        int lmax = 0, rmax = 0;        int value = curNode.val; // 包含当前节点的最大路径和        if(curNode.left != null){            lmax = getMaxSumWithCurNode(curNode.left);        }        if(curNode.right != null){            rmax = getMaxSumWithCurNode(curNode.right);        }                value = value + (lmax>0?lmax:0) + (rmax>0?rmax:0) ;        if(value > maxSum)            maxSum = value;        // 注意这里的返回值，取左右子树其中一条路径            return curNode.val+Math.max( lmax>0?lmax:0, rmax>0?rmax:0 );            }}

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