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:

 1 /**
 2  * Definition for a binary tree node.
 3  * public class TreeNode {
 4  *     int val;
 5  *     TreeNode left;
 6  *     TreeNode right;
 7  *     TreeNode(int x) { val = x; }
 8  * }
 9  */
10 class Solution {
11     int max = 0;
12     public int longestConsecutive(TreeNode root) {
13         dfs(root);
14         return max;
15     }
16     
17     private int[] dfs(TreeNode root){
18         if(root == null){
19             return new int[2];
20         }
21         
22         int inc = 1;
23         int dec = 1;
24         int [] l = dfs(root.left);
25         int [] r = dfs(root.right);
26         if(root.left != null){
27             if(root.left.val - 1 == root.val){
28                 inc = Math.max(inc, l[0]+1);
29             }
30             if(root.left.val + 1 == root.val){
31                 dec = Math.max(dec, l[1]+1);
32             }
33         }
34         
35         if(root.right != null){
36             if(root.right.val - 1 == root.val){
37                 inc = Math.max(inc, r[0]+1);
38             }
39             if(root.right.val + 1 == root.val){
40                 dec = Math.max(dec, r[1]+1);
41             }
42         }
43         
44         max = Math.max(max, dec+inc-1);
45         return new int[]{inc, dec};
46     }
47 }

AC Python:

 1 # Definition for a binary tree node.
 2 # class TreeNode:
 3 #     def __init__(self, val=0, left=None, right=None):
 4 #         self.val = val
 5 #         self.left = left
 6 #         self.right = right
 7 class Solution:
 8     def longestConsecutive(self, root: Optional[TreeNode]) -> int:
 9         self.res = 0
10         self.dfs(root)
11         return self.res
12     
13     def dfs(self, root):
14         if not root:
15             return 0, 0
16         
17         inc, dec = 1, 1
18         lInc, lDec = self.dfs(root.left)
19         rInc, rDec = self.dfs(root.right)
20 
21         if root.left:
22             if root.left.val - 1 == root.val:
23                 inc = max(inc, lInc + 1)
24             if root.left.val + 1 == root.val:
25                 dec = max(dec, lDec + 1)
26         
27         if root.right:
28             if root.right.val - 1 == root.val:
29                 inc = max(inc, rInc + 1)
30             if root.right.val + 1 == root.val:
31                 dec = max(dec, rDec + 1)
32         
33         self.res = max(self.res, inc + dec - 1)
34         return inc, dec

 

posted @ 2018-01-31 07:45  Dylan_Java_NYC  阅读(808)  评论(0)  编辑  收藏  举报