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困难
栈
树
深度优先搜索
二叉搜索树
双指针
二叉树
堆（优先队列）

272. 最接近的二叉搜索树值 II 🔒

English Version

题目描述

给定二叉搜索树的根 root 、一个目标值 target 和一个整数 k ，返回BST中最接近目标的 k 个值。你可以按 任意顺序 返回答案。

题目 保证 该二叉搜索树中只会存在一种 k 个值集合最接近 target

 

示例 1：

输入: root = [4,2,5,1,3]，目标值 = 3.714286，且 k = 2
输出: [4,3]

示例 2:

输入: root = [1], target = 0.000000, k = 1
输出: [1]

 

提示：

  • 二叉树的节点总数为 n
  • 1 <= k <= n <= 104
  • 0 <= Node.val <= 109
  • -109 <= target <= 109

 

进阶：假设该二叉搜索树是平衡的，请问您是否能在小于 O(n)（ n = total nodes ）的时间复杂度内解决该问题呢？

解法

方法一

Python3

# 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)

Java

/**
 * 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);
    }
}

C++

/**
 * 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);
    }
};

Go

/**
 * 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
}