Question

272. Closest Binary Search Tree Value II

中文文档

Description

Given the root of a binary search tree, a target value, and an integer k, return _the _k_ values in the BST that are closest to the_ target. You may return the answer in any order.

You are guaranteed to have only one unique set of k values in the BST that are closest to the target.

 

Example 1:

Input: root = [4,2,5,1,3], target = 3.714286, k = 2
Output: [4,3]

Example 2:

Input: root = [1], target = 0.000000, k = 1
Output: [1]

 

Constraints:

• The number of nodes in the tree is n.
• 1 <= k <= n <= 104.
• 0 <= Node.val <= 109
• -109 <= target <= 109

 

Follow up: Assume that the BST is balanced. Could you solve it in less than O(n) runtime (where n = total nodes)?

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 closestKValues(self, root: TreeNode, target: float, k: int) -> List[int]:
    def dfs(root):
      if root is None:
        return
      dfs(root.left)
      if len(q) < k:
        q.append(root.val)
      else:
        if abs(root.val - target) >= abs(q[0] - target):
          return
        q.popleft()
        q.append(root.val)
      dfs(root.right)

    q = deque()
    dfs(root)
    return list(q)

/**
 * 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> ans;
  private double target;
  private int k;

  public List<Integer> closestKValues(TreeNode root, double target, int k) {
    ans = new LinkedList<>();
    this.target = target;
    this.k = k;
    dfs(root);
    return ans;
  }

  private void dfs(TreeNode root) {
    if (root == null) {
      return;
    }
    dfs(root.left);
    if (ans.size() < k) {
      ans.add(root.val);
    } else {
      if (Math.abs(root.val - target) >= Math.abs(ans.get(0) - target)) {
        return;
      }
      ans.remove(0);
      ans.add(root.val);
    }
    dfs(root.right);
  }
}

/**
 * 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:
  queue<int> q;
  double target;
  int k;

  vector<int> closestKValues(TreeNode* root, double target, int k) {
    this->target = target;
    this->k = k;
    dfs(root);
    vector<int> ans;
    while (!q.empty()) {
      ans.push_back(q.front());
      q.pop();
    }
    return ans;
  }

  void dfs(TreeNode* root) {
    if (!root) return;
    dfs(root->left);
    if (q.size() < k)
      q.push(root->val);
    else {
      if (abs(root->val - target) >= abs(q.front() - target)) return;
      q.pop();
      q.push(root->val);
    }
    dfs(root->right);
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func closestKValues(root *TreeNode, target float64, k int) []int {
  var ans []int
  var dfs func(root *TreeNode)
  dfs = func(root *TreeNode) {
    if root == nil {
      return
    }
    dfs(root.Left)
    if len(ans) < k {
      ans = append(ans, root.Val)
    } else {
      if math.Abs(float64(root.Val)-target) >= math.Abs(float64(ans[0])-target) {
        return
      }
      ans = ans[1:]
      ans = append(ans, root.Val)
    }
    dfs(root.Right)
  }
  dfs(root)
  return ans
}