原题链接在这里：https://leetcode.com/problems/equal-tree-partition/
题目：
Given a binary tree with n nodes, your task is to check if it's possible to partition the tree to two trees which have the equal sum of values after removing exactly one edge on the original tree.
Example 1:
Input:
5
/
10 10
/
2 3
Output: True
Explanation:
5
/
10
Sum: 15
10
/
2 3
Sum: 15
Example 2:
Input:
1
/
2 10
/
2 20
Output: False
Explanation: You can't split the tree into two trees with equal sum after removing exactly one edge on the tree.
Note:
- The range of tree node value is in the range of [-100000, 100000].
- 1 <= n <= 10000
题解：
Calculate sum of each subtree. Check if there is a subtree's sum is half of whole tree.
Use a HashMap to maintain sum value of subtree and its frequency.
Since there is edge case that sum == 0. sum/2 is also 0. Thus there must be more than one occurance of subtree having 0 sum.
Time Complexity: O(n). GetSum() takes O(n).
Space: O(n).
AC Java:
1 /** 2 * Definition for a binary tree node. 3 * public class TreeNode { 4 * int val; 5 * TreeNode left; 6 * TreeNode right; 7 * TreeNode(int x) { val = x; } 8 * } 9 */ 10 class Solution { 11 public boolean checkEqualTree(TreeNode root) { 12 if(root == null){ 13 return true; 14 } 15 16 HashMap<Integer, Integer> hm = new HashMap<Integer, Integer>(); 17 int sum = getSum(root, hm); 18 if(sum == 0){ 19 return hm.get(sum/2) > 1; 20 } 21 22 return sum%2==0 && hm.containsKey(sum/2); 23 } 24 25 private int getSum(TreeNode root, HashMap<Integer, Integer> hm){ 26 if(root == null){ 27 return 0; 28 } 29 30 int sum = getSum(root.left, hm) + root.val + getSum(root.right, hm); 31 hm.put(sum, hm.getOrDefault(sum, 0)+1); 32 return sum; 33 } 34 }