LeetCode 508. Most Frequent Subtree Sum

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原文链接：http://www.cnblogs.com/Dylan-Java-NYC/p/8364876.html

文章标签：

#数据结构与算法 #java

本文介绍了一道LeetCode上的编程题——寻找二叉树中最频繁出现的子树和。通过递归后序遍历的方法，计算每个节点的子树和，并使用哈希表记录每个子树和出现的次数，最终返回出现频率最高的子树和。

原题链接在这里：https://leetcode.com/problems/most-frequent-subtree-sum/description/

题目：

Given the root of a tree, you are asked to find the most frequent subtree sum. The subtree sum of a node is defined as the sum of all the node values formed by the subtree rooted at that node (including the node itself). So what is the most frequent subtree sum value? If there is a tie, return all the values with the highest frequency in any order.

Examples 1
Input:

  5
 /  \
2   -3

return [2, -3, 4], since all the values happen only once, return all of them in any order.

Examples 2
Input:

  5
 /  \
2   -5

return [2], since 2 happens twice, however -5 only occur once.

Note: You may assume the sum of values in any subtree is in the range of 32-bit signed integer.

题解：

自下而上计算每个点的sum, 保存这个sum和对应的count. 更新维护最大的count.

Time Complexity: O(n). Space: O(n). hm size.

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 maxCount;
12     HashMap<Integer, Integer> hm;
13     public int[] findFrequentTreeSum(TreeNode root) {
14         maxCount = 0;
15         hm = new HashMap<Integer, Integer>();
16         
17         postOrder(root);
18         
19         List<Integer> res= new ArrayList<Integer>();
20         for(Map.Entry<Integer, Integer> entry : hm.entrySet()){
21             if(entry.getValue() == maxCount){
22                 res.add(entry.getKey());
23             }
24         }
25         
26         int [] resArr = new int[res.size()];
27         for(int i = 0; i<res.size(); i++){
28             resArr[i] = res.get(i);
29         }
30         
31         return resArr;
32     }
33     
34     private int postOrder(TreeNode root){
35         if(root == null){
36             return 0;
37         }
38         
39         int left = postOrder(root.left);
40         int right = postOrder(root.right);
41         int val = left + right + root.val;
42         hm.put(val, hm.getOrDefault(val, 0)+1);
43         maxCount = Math.max(maxCount, hm.get(val));
44         
45         return val;
46     }
47 }