描述
Given a non-empty binary search tree and a target value, find the value in the BST that is closest to the target.
- 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.
您在真实的面试中是否遇到过这个题?
是
样例
Given root = {1}
, target = 4.428571
, return 1
.
我的第一想法就是先序遍历即可,但是这样做遗漏了最重要的一个条件这是一个二叉搜索树,如果目标比当前节点的值要大,那就没有必要遍历当前节点的左子树了。所以遍历的时候加上筛选条件即可。
/** * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; * } * } */ class Solution { public: /** * @param root: the given BST * @param target: the given target * @return: the value in the BST that is closest to the target */ int closestValue(TreeNode * root, double target) { // write your code here int a=root->val; TreeNode *t=target<a?root->left:root->right; if(!t) return a; int b=closestValue(t,target); return abs(a-target)<abs(b-target)?a:b; } };