LeetCode.508 Most Frequent Subtree Sum_总和最大的subtree-优快云博客

本文介绍了一种算法，用于找到二叉树中最频繁出现的子树和，并返回这些值。通过递归遍历二叉树，计算每个节点的子树和，并使用哈希映射记录每个和的出现次数。

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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.

分析：

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    HashMap<Integer,Integer> hm;
    int maxCount;
    public int[] findFrequentTreeSum(TreeNode root) {
        //给定树结构，求其中各子树的总和出现频率最大的和,如果所有的和都只出现一次，则返回全部，否则返回其频率最高的
        //思路：遍历二叉树，最后将结果记录在全局变量HashMap中，定义一个全局变量记录频率最高的数
        maxCount=0;
        hm=new HashMap<Integer,Integer>();
        
        countSum(root);
        //遍历的hm
        List<Integer> list=new ArrayList<>();
        for(int key:hm.keySet()){
            if(hm.get(key)==maxCount){
                //将同一频率的元素添加list
                list.add(key);
            }
        }
        int [] res=new int[list.size()];
        for(int i=0;i<list.size();i++){
            res[i]=list.get(i);
        }
        return res;
        
    }
    //求二叉树的各个元素的和
    public int countSum(TreeNode t){
        //使用递归实现
        if(t==null)return 0;
       
        int leftSum=countSum(t.left);
        int rightSum=countSum(t.right);
        int curSum=t.val+leftSum+rightSum;
        int times=hm.getOrDefault(curSum,0)+1;
        hm.put(curSum,times);
        //记录频率最高的数
        maxCount=Math.max(times,maxCount);
        return curSum;
    }
}