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.
Solution：
思路：
一次遍历：求出整体sum，并post order向上累计存当前cur_sum 在hashmap中，最后判断hashmap中是否有sum/2即可。
Time Complexity: O(N) Space Complexity: O(N)
Solution Code：
class Solution {
public boolean checkEqualTree(TreeNode root) {
Map<Integer, Integer> map = new HashMap<Integer, Integer>();
int sum = getsum(root, map);
if(sum == 0) return map.getOrDefault(sum, 0) > 1;
return sum % 2 == 0 && map.containsKey(sum / 2);
}
public int getsum(TreeNode root, Map<Integer, Integer> map ){
if(root == null)return 0;
int cur = root.val + getsum(root.left, map) + getsum(root.right, map);
map.put(cur, map.getOrDefault(cur, 0) + 1);
return cur;
}
}
