508. Most Frequent Subtree Sum | 查找子树和中出现次数最多的和

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.

Subscribe to see which companies asked this question.

思路：遍历树，并求和，将和存储在hashmap中，在遍历hashmap。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
   int max = 0;

	public int[] findFrequentTreeSum(TreeNode root) {
		List list = new ArrayList<>();
		HashMap map = new HashMap<>();

		sumOfTree(root, map);
		Iterator iter = map.entrySet().iterator();
		while (iter.hasNext()) {
			Map.Entry entry = (Map.Entry) iter.next();
			int key = (int) entry.getKey();
			int value = (int) entry.getValue();
			if (value == max) {
				list.add(key);
			}

		}
		int[] sums = new int[list.size()];
		for (int i = 0; i < list.size(); i++) {
			sums[i] = list.get(i);
		}
		return sums;
	}

	public int sumOfTree(TreeNode root, HashMap map) {
		if (root == null) {
			return 0;
		}
		int l = sumOfTree(root.left, map);
		int r = sumOfTree(root.right, map);
		map.put(l + r + root.val, map.getOrDefault(l + r + root.val, 0) + 1);
		max = Math.max(max, map.getOrDefault(l + r + root.val, 0));
		return l + r + root.val;
	}
}