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};
}
}