Closest Binary Search Tree Value

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Problem Description

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Solutions

Solution 1: C# (Best: 80 ms)

Metric	Value
Runtime	80 ms
Memory	39.4 MB
Date	2021-11-22

Solution

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int ClosestValue(TreeNode root, double target) {
      var current = root;
    double least = (double) Int32.MaxValue;
    int result = 0;
    while(current != null)
    {
        double newLeast = Math.Abs((double)current.val - target);
        if(newLeast < least)
        {
            result = current.val;
            least = newLeast;
        }
        
        if(target > current.val)
        {
             current = current.right;
        }
        else
        {
            current = current.left;
        }
    }
    
    return result;
    }
}

📜 1 more C# submission(s)

Submission (2021-11-22) — 96 ms, 39.7 MB

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    public int ClosestValue(TreeNode root, double target) {
        if(root.left == null && root.right == null)
    {
        return root.val;
    }
    
    var current = root;
    double least = (double) Int32.MaxValue;
    int currentVal = 0;
    while(current != null)
    {
        double newLeast = Math.Abs((double)current.val - target);
        if(newLeast < least)
        {
            currentVal = current.val;
            least = newLeast;
        }
        
        if(target > current.val)
        {
             current = current.right;
        }
        else
        {
            current = current.left;
        }
    }
    
    return currentVal;
    }
}

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Complexity Analysis

Approach	Time	Space
Solution	To be analyzed	To be analyzed