Code & Func

第17天，又是一道之前没做出来的题目，然而好像并不难啊。

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
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   3
3
  / \
4
 1   4
5
  \
6
   2
7
Output: 1

Example 2:

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Input: root = [5,3,6,2,4,null,null,1], k = 3
2
       5
3
      / \
4
     3   6
5
    / \
6
   2   4
7
  /
8
 1
9
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;
2
int kthSmallest(TreeNode* root, int k) {
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    res = 0;
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    kthSmallestR(root, k);
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    return res;
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}
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bool kthSmallestR(TreeNode *root, int &k) {
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    // cout << root << endl;
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    if (root == nullptr) return false;
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    if (kthSmallestR(root->left, k)) return true;
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    // cout << root->val << endl;
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    if ((--k) == 0) {
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        res = root->val;
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        return true;
3 collapsed lines
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    }
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    return kthSmallestR(root->right, k);
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}

然后我们可以写出非递归版本的：

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int kthSmallest(TreeNode* root, int k) {
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    stack<TreeNode *> st;
3

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    while(root || !st.empty()) {
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        while(root) {
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            st.push(root);
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            root = root->left;
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        }
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        if (st.empty()) break;
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        root = st.top(); st.pop();
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        if (--k == 0) break;
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        root = root->right;
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    }
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    if(root) return root->val;
3 collapsed lines
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    else return -1;
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}

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。