Question

663. Equal Tree Partition

中文文档

Description

Given the root of a binary tree, return true_ if you can partition the tree into two trees with equal sums of values after removing exactly one edge on the original tree_.

 

Example 1:

Input: root = [5,10,10,null,null,2,3]
Output: true

Example 2:

Input: root = [1,2,10,null,null,2,20]
Output: false
Explanation: You cannot split the tree into two trees with equal sums after removing exactly one edge on the tree.

 

Constraints:

• The number of nodes in the tree is in the range [1, 104].
• -105 <= Node.val <= 105

Solutions

Solution 1

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def checkEqualTree(self, root: TreeNode) -> bool:
    def sum(root):
      if root is None:
        return 0
      l, r = sum(root.left), sum(root.right)
      seen.append(l + r + root.val)
      return seen[-1]

    seen = []
    s = sum(root)
    if s % 2 == 1:
      return False
    seen.pop()
    return s // 2 in seen

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {
  private List<Integer> seen;

  public boolean checkEqualTree(TreeNode root) {
    seen = new ArrayList<>();
    int s = sum(root);
    if (s % 2 != 0) {
      return false;
    }
    seen.remove(seen.size() - 1);
    return seen.contains(s / 2);
  }

  private int sum(TreeNode root) {
    if (root == null) {
      return 0;
    }
    int l = sum(root.left);
    int r = sum(root.right);
    int s = l + r + root.val;
    seen.add(s);
    return s;
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  vector<int> seen;

  bool checkEqualTree(TreeNode* root) {
    int s = sum(root);
    if (s % 2 != 0) return false;
    seen.pop_back();
    return count(seen.begin(), seen.end(), s / 2);
  }

  int sum(TreeNode* root) {
    if (!root) return 0;
    int l = sum(root->left), r = sum(root->right);
    int s = l + r + root->val;
    seen.push_back(s);
    return s;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func checkEqualTree(root *TreeNode) bool {
  var seen []int
  var sum func(root *TreeNode) int
  sum = func(root *TreeNode) int {
    if root == nil {
      return 0
    }
    l, r := sum(root.Left), sum(root.Right)
    s := l + r + root.Val
    seen = append(seen, s)
    return s
  }

  s := sum(root)
  if s%2 != 0 {
    return false
  }
  seen = seen[:len(seen)-1]
  for _, v := range seen {
    if v == s/2 {
      return true
    }
  }
  return false
}