# Java - Simple & Easy to understand solution using recursion

• ``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
double min = Double.MAX_VALUE;
double minDiff = Double.MAX_VALUE;
public int closestValue(TreeNode root, double target) {

if(root==null) return 0;

int val = root.val;
if(val == (int)target) return val;

helper(root,target);
return (int)min;
}

private void helper(TreeNode root, double target){
if(root == null) return ;
int val = root.val;
double diff = Math.abs( target - (double)val );
minDiff = Math.min(minDiff, diff );
if(minDiff==diff){
min = root.val;
}
if(root.val > target){ // if root value > target, then go left, if left doesn't exist then root itself would be closest to target, all nodes to the right of root would be greater than root & hence the difference would go on increasing from here.
if(root.left!=null) helper(root.left,target);
else return ;
}else{ // similar explanation as above. If root.val < target then go right, if you go left, then the difference would be more than difference between target and root.val . So there is no point in going to the left hence return;
if(root.right!=null) helper(root.right, target);
else return;
}
}
}
``````

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