Leetcode508 Most Frequent Subtree Sum

Posted 2020-12-04 chason95

研究题意发现，每个结点对应一个Subtree Sum，于是问题就变成了，先traverse一遍，找到每个subtree的sum，之后递归求出subtree的parent结点对应的subtree的sum，可能修改为栈会更快一点？没仔细思考，以后有时间再回来看吧。

    public int[] findFrequentTreeSum(TreeNode root) {
        Set<Integer> ans = new HashSet<>();
        HashMap<Integer,Integer> sums = new HashMap<>();
        getNodeSum(root,sums);
        Set<Map.Entry<Integer,Integer>> set = sums.entrySet();
        int maxnum = 0;
        for(Map.Entry<Integer,Integer> e:set) {
            if(e.getValue()>maxnum) {
                maxnum=e.getValue();
                ans.clear();
                ans.add(e.getKey());
            }
            else if(e.getValue()==maxnum) {
                ans.add(e.getKey());
            }
        }
        int[] result = new int[ans.size()];
        int i=0;
        for(int a:ans) {
            result[i++]=a;
        }
        return result;
    }
    public static int getNodeSum(TreeNode T,HashMap<Integer,Integer> sums){    
        //需要先对T进行判断，这点很重要，不然会报错NullPointerException
        if(T==null) return 0;
        int leftChildSum = T.left!=null?getNodeSum(T.left,sums):0;
        int rightChildSum = T.right!=null?getNodeSum(T.right,sums):0;
        int tempsum = T.val+leftChildSum+rightChildSum;
        if(sums.containsKey(tempsum)) sums.put(tempsum, sums.get(tempsum)+1);
        else sums.put(tempsum, 1);
        return tempsum;
    }