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