270. Closest Binary Search Tree Value - jiejackyzhang/leetcode-note GitHub Wiki

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

Note:

• Given target value is a floating point.
• You are guaranteed to have only one unique value in the BST that is closest to the target.

求BST中跟target最近的数字。我们先设置一个min = root.val，然后用iterative的办法尝试更新min， 然后比较target与root的大小，进行二分查找。

Time Complexity - O(logn)， Space Complexity - O(1)。

##Approach 1: iterative

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int closestValue(TreeNode root, double target) {
        if (root == null) return 0;
        int min = root.val;
        while(root != null) {
            min = Math.abs(target - root.val) < Math.abs(target - min) ? root.val : min;
            root = root.val < target ? root.right : root.left;
        }
        
        return min;
    }
}

##Approach 2: recursive

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public int closestValue(TreeNode root, double target) {
        if (root == null) return 0;
        TreeNode child = target < root.val ? root.left : root.right;
        if (child == null) {
            return root.val; 
        }
        int childClosest = closestValue(child, target);
        return Math.abs(root.val - target) < Math.abs(childClosest - target) ? root.val : childClosest;
    }
}