663 Equal Tree Partition

663. Equal Tree Partition

1. Question

Given a binary tree withnnodes, 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:

1. The range of tree node value is in the range of [-100000, 100000].

2. 1 <= n <= 10000

2. Implementation

(1) DFS

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public boolean checkEqualTree(TreeNode root) {
        Map<Integer, Integer> map = new HashMap();
        int sum = getTreeSum(root, map);

        if (sum == 0) {
            return map.getOrDefault(sum, 0) >= 2;
        }

        return sum % 2 == 0 ? map.containsKey(sum / 2) : false;
    }

    public int getTreeSum(TreeNode node, Map<Integer, Integer> map) {
        if (node == null) {
            return 0;
        }

        int leftSum = getTreeSum(node.left, map);
        int rightSum = getTreeSum(node.right, map);

        int sum = node.val + leftSum + rightSum;
        map.put(sum, map.getOrDefault(sum, 0) + 1);
        return sum;
    }
}

3. Time & Space Complexity

DFS: 时间复杂度O(n), 空间复杂度O(k), k是map里不同sum的个数