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_508.java
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package com.fishercoder.solutions.firstthousand;
import com.fishercoder.common.classes.TreeNode;
import java.util.ArrayList;
import java.util.Collections;
import java.util.HashMap;
import java.util.LinkedList;
import java.util.List;
import java.util.Map;
public class _508 {
public static class Solution1 {
// my purely original but verbose solution
public int[] findFrequentTreeSum(TreeNode root) {
if (root == null) {
return new int[] {};
}
Map<TreeNode, Integer> map = new HashMap();
postOrder(root, map);
Map<Integer, Integer> frequencyMap = new HashMap<>();
for (Map.Entry entry : map.entrySet()) {
frequencyMap.put(
(Integer) entry.getValue(),
frequencyMap.getOrDefault(entry.getValue(), 0) + 1);
}
List<Map.Entry<Integer, Integer>> list = new LinkedList<>(frequencyMap.entrySet());
Collections.sort(list, (o1, o2) -> (o2.getValue()).compareTo(o1.getValue()));
int mostFrequency = list.get(0).getValue();
List<Integer> topFrequencyList = new ArrayList<>();
topFrequencyList.add(list.get(0).getKey());
int i = 1;
while (i < list.size() && list.get(i).getValue() == mostFrequency) {
topFrequencyList.add(list.get(i).getKey());
i++;
}
int[] result = new int[topFrequencyList.size()];
for (int j = 0; j < topFrequencyList.size(); j++) {
result[j] = topFrequencyList.get(j);
}
return result;
}
private int postOrder(TreeNode root, Map<TreeNode, Integer> map) {
int left = 0;
int right = 0;
if (root.left != null) {
left = postOrder(root.left, map);
}
if (root.right != null) {
right = postOrder(root.right, map);
}
if (root.left == null && root.right == null) {
map.put(root, root.val);
return root.val;
}
int sum = left + right + root.val;
map.put(root, sum);
return sum;
}
}
public static class Solution2 {
// my 2nd purely original but verbose solution
public int[] findFrequentTreeSum(TreeNode root) {
Map<Integer, Integer> map = new HashMap<>();
dfs(root, map);
List<Map.Entry<Integer, Integer>> entryList = new ArrayList<>(map.entrySet());
Collections.sort(entryList, (a, b) -> b.getValue() - a.getValue());
List<Integer> list = new ArrayList<>();
for (int i = 0; i < entryList.size(); i++) {
if (list.size() == 0) {
list.add(entryList.get(i).getKey());
} else {
if (map.get(list.get(0)) == entryList.get(i).getValue()) {
list.add(entryList.get(i).getKey());
} else {
break;
}
}
}
int[] result = new int[list.size()];
for (int i = 0; i < list.size(); i++) {
result[i] = list.get(i);
}
return result;
}
private int dfs(TreeNode root, Map<Integer, Integer> map) {
if (root == null) {
return 0;
}
if (root.left == null && root.right == null) {
map.put(root.val, map.getOrDefault(root.val, 0) + 1);
return root.val;
}
int leftVal = 0;
if (root.left != null) {
leftVal = dfs(root.left, map);
}
int rightVal = 0;
if (root.right != null) {
rightVal = dfs(root.right, map);
}
int subtreeSum = leftVal + rightVal + root.val;
map.put(subtreeSum, map.getOrDefault(subtreeSum, 0) + 1);
return subtreeSum;
}
}
public static class Solution3 {
/*
* Use post-order traversal for this problem as it needs to process subtree first before processing the root.
*/
Map<Integer, Integer> map = new HashMap<>();
public int[] findFrequentTreeSum(TreeNode root) {
postOrder(root);
int mostFreq = -1;
for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
mostFreq = Math.max(mostFreq, entry.getValue());
}
List<Integer> list = new ArrayList<>();
for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
if (entry.getValue() == mostFreq) {
list.add(entry.getKey());
}
}
return list.stream().mapToInt(integer -> integer).toArray();
}
private int postOrder(TreeNode root) {
if (root == null) {
return 0;
}
int leftSum = postOrder(root.left);
int rightSum = postOrder(root.right);
int subtreeSum = leftSum + rightSum + root.val;
map.put(subtreeSum, map.getOrDefault(subtreeSum, 0) + 1);
return subtreeSum;
}
}
// a more concise and space-efficient solution:
// https://discuss.leetcode.com/topic/77775/verbose-java-solution-postorder-traverse-hashmap-18ms
// the key difference between the above post and my original solution is that it's using
// Frequency as the key of the HashMap
}