[LeetCode] 270. Closest Binary Search Tree Value
grandyang opened this issue · comments
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.
Example:
Input: root = [4,2,5,1,3], target = 3.714286
4
/ \
2 5
/ \
1 3
Output: 4
这道题让我们找一个二分搜索数的跟给定值最接近的一个节点值,由于是二分搜索树,所以博主最先想到用中序遍历来做,一个一个的比较,维护一个最小值,不停的更新,实际上这种方法并没有提高效率,用其他的遍历方法也可以,参见代码如下:
解法一:
class Solution {
public:
int closestValue(TreeNode* root, double target) {
double d = numeric_limits<double>::max();
int res = 0;
stack<TreeNode*> s;
TreeNode *p = root;
while (p || !s.empty()) {
while (p) {
s.push(p);
p = p->left;
}
p = s.top(); s.pop();
if (d >= abs(target - p->val)) {
d = abs(target - p->val);
res = p->val;
}
p = p->right;
}
return res;
}
};
实际我们可以利用二分搜索树的特点 (左<根<右) 来快速定位,由于根节点是中间值,在往下遍历时,根据目标值和根节点的值大小关系来比较,如果目标值小于节点值,则应该找更小的值,于是到左子树去找,反之去右子树找,参见代码如下:
解法二:
class Solution {
public:
int closestValue(TreeNode* root, double target) {
int res = root->val;
while (root) {
if (abs(res - target) >= abs(root->val - target)) {
res = root->val;
}
root = target < root->val ? root->left : root->right;
}
return res;
}
};
以上两种方法都是迭代的方法,下面来看递归的写法,下面这种递归的写法和上面迭代的方法思路相同,都是根据二分搜索树的性质来优化查找,但是递归的写法用的是回溯法,先遍历到叶节点,然后一层一层的往回走,把最小值一层一层的运回来,参见代码如下:
解法三:
class Solution {
public:
int closestValue(TreeNode* root, double target) {
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;
}
};
再来看另一种递归的写法,思路和上面的都相同,写法上略有不同,用if来分情况,参见代码如下:
解法三:
class Solution {
public:
int closestValue(TreeNode* root, double target) {
int res = root->val;
if (target < root->val && root->left) {
int l = closestValue(root->left, target);
if (abs(res - target) >= abs(l - target)) res = l;
} else if (target > root->val && root->right) {
int r = closestValue(root->right, target);
if (abs(res - target) >= abs(r - target)) res = r;
}
return res;
}
};
最后来看一种分治法的写法,这种方法相当于解法一的递归写法,并没有利用到二分搜索树的性质来优化搜索,参见代码如下:
解法四:
class Solution {
public:
int closestValue(TreeNode* root, double target) {
double diff = numeric_limits<double>::max();
int res = 0;
helper(root, target, diff, res);
return res;
}
void helper(TreeNode *root, double target, double &diff, int &res) {
if (!root) return;
if (diff >= abs(root->val - target)) {
diff = abs(root->val - target);
res = root->val;
}
helper(root->left, target, diff, res);
helper(root->right, target, diff, res);
}
};
Github 同步地址:
类似题目:
Closest Binary Search Tree Value II
Search in a Binary Search Tree
参考资料:
https://leetcode.com/problems/closest-binary-search-tree-value/