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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,
1
/ \
2 3
SOLUTION 1:
计算树的最长path有2种情况：
1. 通过根的path.
(1)如果左子树从左树根到任何一个Node的path大于零，可以链到root上
(2)如果右子树从右树根到任何一个Node的path大于零，可以链到root上
2. 不通过根的path. 这个可以取左子树及右子树的path的最大值。
所以创建一个inner class:
记录2个值：
1. 本树的最大path。
2. 本树从根节点出发到任何一个节点的最大path.
注意，当root == null,以上2个值都要置为Integer_MIN_VALUE; 因为没有节点可取的时候，是不存在solution的。以免干扰递归的计算
1 /** 2 * Definition for binary tree 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 public class Solution { 11 public class ReturnType { 12 int maxSingle; 13 int max; 14 ReturnType (int maxSingle, int max) { 15 this.max = max; 16 this.maxSingle = maxSingle; 17 } 18 } 19 20 public int maxPathSum(TreeNode root) { 21 return dfs(root).max; 22 } 23 24 public ReturnType dfs(TreeNode root) { 25 ReturnType ret = new ReturnType(Integer.MIN_VALUE, Integer.MIN_VALUE); 26 if (root == null) { 27 return ret; 28 } 29 30 ReturnType left = dfs(root.left); 31 ReturnType right = dfs(root.right); 32 33 int cross = root.val; 34 35 // if any of the path of left and right is below 0, don‘t add it. 36 cross += Math.max(0, left.maxSingle); 37 cross += Math.max(0, right.maxSingle); 38 39 // 注意，这里不可以把Math.max(left.maxSingle, right.maxSingle) 与root.val加起来， 40 // 会有可能越界！ 41 int maxSingle = Math.max(left.maxSingle, right.maxSingle); 42 43 // may left.maxSingle and right.maxSingle are below 0 44 maxSingle = Math.max(maxSingle, 0); 45 maxSingle += root.val; 46 47 ret.maxSingle = maxSingle; 48 ret.max = Math.max(right.max, left.max); 49 ret.max = Math.max(ret.max, cross); 50 51 return ret; 52 } 53 }
GITHUB:
https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/MaxPathSum.java
LeetCode: Binary Tree Maximum Path Sum 解题报告
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原文地址：http://www.cnblogs.com/yuzhangcmu/p/4172855.html
