Leetcode-549. Binary Tree Longest Consecutive Sequence II 算法

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。
博客链接：mcf171的博客

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Given a binary tree, you need to find the length of Longest Consecutive Path in Binary Tree.
Especially, this path can be either increasing or decreasing. For example, [1,2,3,4] and [4,3,2,1] are both considered valid, but the path [1,2,4,3] is not valid. On the other hand, the path can be in the child-Parent-child order, where not necessarily be parent-child order.
Example 1:

Input: 1 / \ 23 Output: 2 Explanation: The longest consecutive path is [1, 2] or [2, 1].

Example 2:

Input: 2 / \ 13 Output: 3 Explanation: The longest consecutive path is [1, 2, 3] or [3, 2, 1].

这个题目还是蛮简单的，有一个需要注意的地方就是必须是连续值。

public class Solution { private int max = 1; public int longestConsecutive(TreeNode root) { if(root == null) return 0; help(root); return max; }public int[] help(TreeNode root){ if(root == null) return null; int[] right = help(root.right); int[] left = help(root.left); int[] result = new int[]{1,1}; if(right == null && left == null) return result; if(root.right != null){ if(root.val == root.right.val + 1) { max = Math.max(max, right[0] + 1); result[0] = right[0] + 1; result[1] = 1; }else if(root.val == root.right.val - 1){ max = Math.max(max, right[1] + 1); result[0] = 1; result[1] = right[1] + 1; } } if(root.left != null){ if(root.val == root.left.val + 1) { max = Math.max(max, left[0] + result[1]); result[0] = Math.max(result[0], left[0] + 1); }else if(root.val == root.left.val - 1){ max = Math.max(max, left[1] + result[0]); result[1] = Math.max(result[1], left[1] + 1); } }return result; } }

【Leetcode-549. Binary Tree Longest Consecutive Sequence II】