Code & Fun

第17天,又是一道之前没做出来的题目,然而好像并不难啊。

今天的题目是 Kth Smallest Element in a BST :


Given a binary search tree, write a function kthSmallest to find the kth smallest element in it.

Note:
You may assume k is always valid, 1 ≤ k ≤ BST’s total elements.

Example 1:

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Input: root = [3,1,4,null,2], k = 1
3
/ \
1 4
\
2
Output: 1

Example 2:

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Input: root = [5,3,6,2,4,null,null,1], k = 3
5
/ \
3 6
/ \
2 4
/
1
Output: 3

Follow up:
What if the BST is modified (insert/delete operations) often and you need to find the kth smallest frequently? How would you optimize the kthSmallest routine?


题意很简单就是求BST中第k小的数字,然后BST本身就包含一定的顺序信息,利用BST中序遍历是有序的性质,我们可以很快的把这道题写出来:

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int res;
int kthSmallest(TreeNode* root, int k) {
res = 0;
kthSmallestR(root, k);
return res;
}
bool kthSmallestR(TreeNode *root, int &k) {
// cout << root << endl;
if (root == nullptr) return false;

if (kthSmallestR(root->left, k)) return true;
// cout << root->val << endl;
if ((--k) == 0) {
res = root->val;
return true;
}
return kthSmallestR(root->right, k);
}

然后我们可以写出非递归版本的:

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int kthSmallest(TreeNode* root, int k) {
stack<TreeNode *> st;

while(root || !st.empty()) {
while(root) {
st.push(root);
root = root->left;
}
if (st.empty()) break;
root = st.top(); st.pop();
if (--k == 0) break;
root = root->right;
}

if(root) return root->val;
else return -1;

}

BTW,这道题的测试有点不稳定,同一个代码会测试出不同的时间。


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