LeetCode 题解(249) : Closest Binary Search Tree Value

来源：互联网发布：苹果6手机壳淘宝编辑：程序博客网时间：2024/05/01 08:40

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.

题解：

递归。用全局变量保存当前最接近的节点值。

C++版：

/** * Definition for a binary tree node. * struct TreeNode { *     int val; *     TreeNode *left; *     TreeNode *right; *     TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public:    int closestValue(TreeNode* root, double target) {        if(root->val == target)            return target;                double cur = abs(root->val - target);            if(cur < curM) {            curM = cur;            rootV = root->val;        }        if(root->val < target) {            if(root->right != NULL)                return closestValue(root->right, target);            else                return rootV;        } else {            if(root->left != NULL)                return closestValue(root->left, target);            else                return rootV;        }    }    private:    double rootV = INT_MAX;    double curM = std::numeric_limits<double>::max();};

Java版：

/** * 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.val == target)            return root.val;                    double cur = Math.abs(root.val - target);        if(cur < curM) {            rootV = root.val;            curM = cur;        }        if(root.val > target) {            if(root.left != null)                return closestValue(root.left, target);            else                return rootV;        } else {            if(root.right != null)                return closestValue(root.right, target);            else                return rootV;        }    }        private int rootV = Integer.MAX_VALUE;    private double curM = Double.MAX_VALUE;}

Python版：

# Definition for a binary tree node.# class TreeNode(object):#     def __init__(self, x):#         self.val = x#         self.left = None#         self.right = Noneimport sysclass Solution(object):    def __init__(self):        self.v = sys.maxint        self.curM = sys.maxint        def closestValue(self, root, target):        """        :type root: TreeNode        :type target: float        :rtype: int        """        if root.val == target:            return root.val        cur = abs(root.val - target)        if cur < self.curM:            self.curM = cur            self.v = root.val        if root.val > target:            if root.left != None:                return self.closestValue(root.left, target)            else:                return self.v        else:            if root.right != None:                return self.closestValue(root.right, target)            else:                return self.v

0 0