LeetCode 549. Binary Tree Longest Consecutive Sequence II

原题链接在这里：https://leetcode.com/problems/binary-tree-longest-consecutive-sequence-ii/description/

题目：

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

Example 2:

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

Note: All the values of tree nodes are in the range of [-1e7, 1e7].

题解：

与Binary Tree Longest Consecutive Sequence类似.

采用bottom-up的方法dfs. 每个点同时维护能向下延展的最大increasing 和 decreasing长度.

Time Complexity: O(n). Space: O(logn).

AC Java:

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 class Solution {
     int max = 0;
     public int longestConsecutive(TreeNode root) {
         dfs(root);
         return max;
     }

     private int[] dfs(TreeNode root){
         if(root == null){
             return new int[2];
         }

         int inc = 1;
         int dec = 1;
         int [] l = dfs(root.left);
         int [] r = dfs(root.right);
         if(root.left != null){
             if(root.left.val - 1 == root.val){
                 inc = Math.max(inc, l[0]+1);
             }
             if(root.left.val + 1 == root.val){
                 dec = Math.max(dec, l[1]+1);
             }
         }

         if(root.right != null){
             if(root.right.val - 1 == root.val){
                 inc = Math.max(inc, r[0]+1);
             }
             if(root.right.val + 1 == root.val){
                 dec = Math.max(dec, r[1]+1);
             }
         }

         max = Math.max(max, dec+inc-1);
         return new int[]{inc, dec};
     }
 }