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