270. Closest Binary Search Tree Value

题目：

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.

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

题解：

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)。

/**
 * 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) {
        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;
    }
}

二刷:

这道题也是主要考察binary search。方法和一刷一样。  可以有递归和迭代。

Java:

迭代

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

/**
 * 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) {
        int min = root.val;
        while (root != null) {
            min = Math.abs(root.val - target) < Math.abs(min -target) ? root.val : min;
            root = target < root.val ? root.left : root.right;
        }
        return min;
    }
}

递归:

/**
 * 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) {
        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;
    }
}

三刷:

Java:

/**
 * 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(root.val - target) < Math.abs(min - target) ? root.val : min);
            root = (root.val < target) ? root.right : root.left;
        }
        return min;
    }
}

 

Reference:

https://leetcode.com/discuss/54438/4-7-lines-recursive-iterative-ruby-c-java-python