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Formatted question description: https://leetcode.ca/all/2476.html

2476. Closest Nodes Queries in a Binary Search Tree

  • Difficulty: Medium.
  • Related Topics: .
  • Similar Questions: Closest Binary Search Tree Value, Closest Binary Search Tree Value II, Search in a Binary Search Tree.

Problem

You are given the root of a **binary search tree **and an array queries of size n consisting of positive integers.

Find a 2D array answer of size n where answer[i] = [mini, maxi]:

  • mini is the largest value in the tree that is smaller than or equal to queries[i]. If a such value does not exist, add -1 instead.

  • maxi is the smallest value in the tree that is greater than or equal to queries[i]. If a such value does not exist, add -1 instead.

Return the array answer.

  Example 1:

Input: root = [6,2,13,1,4,9,15,null,null,null,null,null,null,14], queries = [2,5,16]
Output: [[2,2],[4,6],[15,-1]]
Explanation: We answer the queries in the following way:
- The largest number that is smaller or equal than 2 in the tree is 2, and the smallest number that is greater or equal than 2 is still 2. So the answer for the first query is [2,2].
- The largest number that is smaller or equal than 5 in the tree is 4, and the smallest number that is greater or equal than 5 is 6. So the answer for the second query is [4,6].
- The largest number that is smaller or equal than 16 in the tree is 15, and the smallest number that is greater or equal than 16 does not exist. So the answer for the third query is [15,-1].

Example 2:

Input: root = [4,null,9], queries = [3]
Output: [[-1,4]]
Explanation: The largest number that is smaller or equal to 3 in the tree does not exist, and the smallest number that is greater or equal to 3 is 4. So the answer for the query is [-1,4].

  Constraints:

  • The number of nodes in the tree is in the range [2, 105].

  • 1 <= Node.val <= 106

  • n == queries.length

  • 1 <= n <= 105

  • 1 <= queries[i] <= 106

Solution (Java, C++, Python)

  • /**
 * 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> nums = new ArrayList<>();

    public List<List<Integer>> closestNodes(TreeNode root, List<Integer> queries) {
        dfs(root);
        List<List<Integer>> ans = new ArrayList<>();
        for (int v : queries) {
            int i = search(v + 1) - 1;
            int j = search(v);
            int mi = i >= 0 && i < nums.size() ? nums.get(i) : -1;
            int mx = j >= 0 && j < nums.size() ? nums.get(j) : -1;
            ans.add(Arrays.asList(mi, mx));
        }
        return ans;
    }

    private int search(int x) {
        int left = 0, right = nums.size();
        while (left < right) {
            int mid = (left + right) >> 1;
            if (nums.get(mid) >= x) {
                right = mid;
            } else {
                left = mid + 1;
            }
        }
        return left;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        nums.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:
    vector<vector<int>> closestNodes(TreeNode* root, vector<int>& queries) {
        vector<int> nums;
        function<void(TreeNode* root)> dfs = [&](TreeNode* root) {
            if (!root) return;
            dfs(root->left);
            nums.emplace_back(root->val);
            dfs(root->right);
        };
        dfs(root);
        vector<vector<int>> ans;
        int n = nums.size();
        for (int& v : queries) {
            int i = upper_bound(nums.begin(), nums.end(), v) - nums.begin() - 1;
            int j = lower_bound(nums.begin(), nums.end(), v) - nums.begin();
            int mi = i >= 0 && i < n ? nums[i] : -1;
            int mx = j >= 0 && j < n ? nums[j] : -1;
            ans.push_back({mi, mx});
        }
        return ans;
    }
};

  • # 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 closestNodes(
        self, root: Optional[TreeNode], queries: List[int]
    ) -> List[List[int]]:
        def dfs(root):
            if root is None:
                return
            dfs(root.left)
            nums.append(root.val)
            dfs(root.right)

        nums = []
        dfs(root)
        ans = []
        for v in queries:
            i = bisect_right(nums, v) - 1
            j = bisect_left(nums, v)
            mi = nums[i] if 0 <= i < len(nums) else -1
            mx = nums[j] if 0 <= j < len(nums) else -1
            ans.append([mi, mx])
        return ans


  • /**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func closestNodes(root *TreeNode, queries []int) (ans [][]int) {
	nums := []int{}
	var dfs func(*TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		dfs(root.Left)
		nums = append(nums, root.Val)
		dfs(root.Right)
	}
	dfs(root)
	n := len(nums)
	for _, v := range queries {
		i := sort.SearchInts(nums, v+1) - 1
		j := sort.SearchInts(nums, v)
		mi, mx := -1, -1
		if i >= 0 && i < n {
			mi = nums[i]
		}
		if j >= 0 && j < n {
			mx = nums[j]
		}
		ans = append(ans, []int{mi, mx})
	}
	return
}

  • /**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function closestNodes(root: TreeNode | null, queries: number[]): number[][] {
    const nums: number[] = [];
    const dfs = (root: TreeNode | null) => {
        if (!root) {
            return;
        }
        dfs(root.left);
        nums.push(root.val);
        dfs(root.right);
    };
    const search = (x: number): number => {
        let [l, r] = [0, nums.length];
        while (l < r) {
            const mid = (l + r) >> 1;
            if (nums[mid] >= x) {
                r = mid;
            } else {
                l = mid + 1;
            }
        }
        return l;
    };
    dfs(root);
    const ans: number[][] = [];
    for (const x of queries) {
        const i = search(x + 1) - 1;
        const j = search(x);
        const mi = i >= 0 ? nums[i] : -1;
        const mx = j < nums.length ? nums[j] : -1;
        ans.push([mi, mx]);
    }
    return ans;
}

Explain:

nope.

Complexity:

  • Time complexity : O(n).
  • Space complexity : O(n).

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