1、问题描述
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 must contain at least one node and does not need to go through the root.
For example:
Given the below binary tree,
1
/ 2 3
Return 6.
2、边界条件：无
3、思路：最大路径值，一个单向路径的最大值已经解决了，一个节点向下的左右两条路径。这道题目可以跨一个节点的两个子节点。这里需要多考虑一个路径，就是左右路径和该节点组成的长路径。另外一点需要注意，一条路径如果<0 ,则可以不连接。需要一个变量来传递当前的最大路径值。
4、代码实现
/** * Definition for a binary tree node. * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ class Solution { public int maxPathSum(TreeNode root) { int[] maxVal = new int[1]; maxVal[0] = Integer.MIN_VALUE; maxPathSum(root, maxVal); return maxVal[0]; } public int maxPathSum(TreeNode root, int[] maxVal) { if (root == null) { return 0; } int leftMaxVal = Math.max(0, maxPathSum(root.left, maxVal)); int rightMaxVal = Math.max(0, maxPathSum(root.right, maxVal)); maxVal[0] = Math.max(maxVal[0], leftMaxVal + rightMaxVal + root.val); return Math.max(leftMaxVal, rightMaxVal) + root.val; } }
5、时间复杂度：O(N), 空间复杂度：
6.api：
leetcode--124. Binary Tree Maximum Path Sum
原文：http://www.cnblogs.com/shihuvini/p/7465510.html