508 Most Frequent Subtree Sum¶
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:
Note: You may assume the sum of values in any subtree is in the range of 32-bit signed integer.
Solution¶
- hasmap
- subtree
- sum
public int[] findFrequentTreeSum(TreeNode root) {
if (root == null) {
return new int[0];
}
HashMap<Integer, Integer> hashmap = new HashMap<Integer, Integer>();
helper(root, hashmap);
int max = 0;
ArrayList<Integer> ans = new ArrayList<Integer>();
for (Integer count : hashmap.values()) {
max = Math.max(max, count);
}
for (Map.Entry<Integer, Integer> entry : hashmap.entrySet()) {
if (entry.getValue() == max) {
ans.add(entry.getKey());
}
}
int[] res = new int[ans.size()];
for (int i = 0; i < res.length; i++) {
res[i] = ans.get(i);
}
return res;
}
private int helper(TreeNode root, HashMap<Integer, Integer> hashmap){
if (root == null) {
return 0;
}
int sum = root.val + helper(root.left, hashmap) + helper(root.right, hashmap);
hashmap.put(sum, hashmap.containsKey(sum) ? hashmap.get(sum) + 1 : 1);
return sum;
}
c++
// Running time: 18 ms
class Solution {
public:
vector<int> findFrequentTreeSum(TreeNode* root) {
unordered_map<int, int> freqs;
int max_freq = -1;
vector<int> ans;
(void)treeSum(root, freqs, max_freq, ans);
return ans;
}
private:
int treeSum(TreeNode* root, unordered_map<int, int>& freqs, int& max_freq, vector<int>& ans) {
if (!root) return 0;
int sum = root->val +
treeSum(root->left, freqs, max_freq, ans) +
treeSum(root->right, freqs, max_freq, ans);
int freq = ++freqs[sum];
if (freq > max_freq) {
max_freq = freq;
ans.clear();
}
if (freq == max_freq)
ans.push_back(sum);
return sum;
}
};