LeetCode – Closest Binary Search Tree Value (Java)

Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target.

Java Solution 1 - Recursion

Recursively traverse down the root. When target is less than root, go left; when target is greater than root, go right.

public class Solution {
    int goal;
    double min = Double.MAX_VALUE;
 
    public int closestValue(TreeNode root, double target) {
        helper(root, target);
        return goal;
    }
 
    public void helper(TreeNode root, double target){
        if(root==null)
            return;
 
        if(Math.abs(root.val - target) < min){
            min = Math.abs(root.val-target);
            goal = root.val;
        } 
 
        if(target < root.val){
            helper(root.left, target);
        }else{
            helper(root.right, target);
        }
    }
}

Java Solution 2 - Iteration

public int closestValue(TreeNode root, double target) {
    double min=Double.MAX_VALUE;
    int result = root.val;
 
    while(root!=null){
        if(target>root.val){
 
            double diff = Math.abs(root.val-target);
            if(diff<min){
                min = Math.min(min, diff);
                result = root.val;
            }
            root = root.right;
        }else if(target<root.val){
 
            double diff = Math.abs(root.val-target);
            if(diff<min){
                min = Math.min(min, diff);
                result = root.val;
            }
            root = root.left;
        }else{
            return root.val;
        }
    }
 
    return result;
}
Category >> Algorithms  
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  • Harshad Sathe

    It finds the difference between the root value and target(ignores -ve sign) and basically remembers the min value which is “right now” closest to the target value in every recursive call

  • Bhanu

    I couldnt understand the following part of the code. can anyone explain

    if(Math.abs(root.val – target) < min){

    min = Math.abs(root.val-target);

    goal = root.val;

    }