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};
    }
}