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  1. Trees
  2. Closest Binary Search Tree Value

Trees

Closest Binary Search Tree Value

Problem Description

Given the root of a binary search tree and a target value, return the value in the BST that is closest to the target.

Constraints:

  • The number of nodes in the tree is in the range [1, 10**4].
  • 0 <= Node.val <= 10**9
  • -10**9 <= target <= 10**9

https://leetcode.com/problems/closest-binary-search-tree-value/

Examples

Example 1
Input: root = [4,2,5,1,3], target = 3.714286
Output: 4
Explanation:
       4
     /   \
   2      5
 /   \
1     3
Example 2
Input: root = [1], target = 4.428571
Output: 1

How to Solve

The tree traversal can be iterative or recursive. It is a binary search tree, so looking the given target value, go deeper left or right. The solution here took an iterative approach. Stating from root node value, if a difference of current is smaller than previous, update the result value. Then go left if the target is smaller than current node value.

Solution

  • /**
     * 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 ClosestBinarySearchTreeValue {
    public:
        int closestValue(TreeNode* root, double target) {
            int result = root->val;
            while (root) {
                if (abs(root->val - target) < abs(result - target)) {
                    result = root->val;
                }
                if (root->val > target) {
                    root = root->left;
                } else {
                    root = root->right;
                }
            }
            return result;
        }
    };
    
  • /**
     * 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 {
        public int closestValue(TreeNode root, double target) {
            int result = root.val;
            while (root != null) {
                if (Math.abs(root.val - target) < Math.abs(result - target)) {
                    result = root.val;
                }
                if (root.val > target) {
                    root = root.left;
                } else {
                    root = root.right;
                }
            }
            return result;
        }
    }
    
  • /**
     * Definition for a binary tree node.
     * function TreeNode(val, left, right) {
     *     this.val = (val===undefined ? 0 : val)
     *     this.left = (left===undefined ? null : left)
     *     this.right = (right===undefined ? null : right)
     * }
     */
    /**
     * @param {TreeNode} root
     * @param {number} target
     * @return {number}
     */
    var closestValue = function(root, target) {
        let result = root.val;
        while (root) {
            if (Math.abs(root.val - target) < Math.abs(result - target)) {
                result = root.val;
            }
            if (root.val > target) {
                root = root.left;
            } else {
                root = root.right;
            }
        }
        return result;
    };
    
  • # 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 ClosestBinarySearchTreeValue:
        def closestValue(self, root: Optional[TreeNode], target: float) -> int:
            result = root.val
            while root:
                if abs(root.val - target) < abs(result - target):
                    result = root.val
                if root.val > target:
                    root = root.left
                else:
                    root = root.right
            return result
    
  • # Definition for a binary tree node.
    # class TreeNode
    #     attr_accessor :val, :left, :right
    #     def initialize(val = 0, left = nil, right = nil)
    #         @val = val
    #         @left = left
    #         @right = right
    #     end
    # end
    # @param {TreeNode} root
    # @param {Float} target
    # @return {Integer}
    def closest_value(root, target)
        result = root.val
        while root do
            if (root.val - target).abs < (result - target).abs
                result = root.val
            end
            if root.val > target
                root = root.left
            else
                root = root.right
            end
        end
        result
    end
    

Complexities

  • Time: O(log(n))
  • Space: O(1)
Easy
Binary Search Tree
Depth-First Search