leetcode解题: Most Frequent Subtree Sum (508)

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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.

解法1：Tree traversal + HashMap O(N)

计算每一个节点的sum很方便，就是leftsum + rightsum。然后用一个hashmap维护每一个sum的出现的频率。
最后统计最大频率的key。下面的解法我用了排序，实际上不需要排序。只需要遍历两遍，一遍找出最大频率，第二部就是找出所有频率为max的key就可以了。
Java

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int[] findFrequentTreeSum(TreeNode root) {
        if (root == null) {
            return new int[0];
        }
        
        HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
        helper(root, map);
        
        // Sort map entry based on the frequency
        Comparator<Map.Entry<Integer, Integer>> comparator = new Comparator<Map.Entry<Integer, Integer>>() {
            public int compare(Map.Entry<Integer, Integer> left, Map.Entry<Integer, Integer> right) {
                return right.getValue().compareTo(left.getValue());
            }
        };
        List<Map.Entry<Integer, Integer>> res = new ArrayList<Map.Entry<Integer, Integer>>();
        for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
            res.add(entry);
        }
        
        res.sort(comparator);   // Sort by frequency in decreasing order
        int maxFreq = res.get(0).getValue();
        List<Integer> temp = new ArrayList<Integer>();
        for (int i = 0; i < res.size(); ++i) {
             if (res.get(i).getValue() == maxFreq) {
                 temp.add(res.get(i).getKey());
             }
        }
        
        int[] resArray = new int[temp.size()];
        for (int i = 0; i < temp.size(); ++i) {
            resArray[i] = temp.get(i);
        }
        
        return resArray;
    }
    
    private int helper(TreeNode root, Map<Integer, Integer> freqMap) {
        if (root == null) {
            return 0;
        }
        
        int left = helper(root.left, freqMap);
        int right = helper(root.right, freqMap);
        
        int sum = left + right + root.val;
        freqMap.put(sum, freqMap.getOrDefault(sum, 0) + 1);
        return sum;
    }
}