題目鏈接
tag:

• Medium；
• Stack；

question：
??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:

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

Example 2:

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?

思路：
??這道題給的提示是讓我們用BST的性質(zhì)來解題，最重要的性質(zhì)是就是左<根<右，那么如果用中序遍歷所有的節(jié)點(diǎn)就會得到一個(gè)有序數(shù)組。所以解題的關(guān)鍵還是中序遍歷啊。關(guān)于二叉樹的中序遍歷可以參見我之前的博客Binary Tree Inorder Traversal 二叉樹中序遍歷，我們來看一種非遞歸的方法，中序遍歷最先遍歷到的是最小的結(jié)點(diǎn)，那么我們只要用一個(gè)計(jì)數(shù)器，每遍歷一個(gè)結(jié)點(diǎn)，計(jì)數(shù)器自增1，當(dāng)計(jì)數(shù)器到達(dá)k時(shí)，返回當(dāng)前結(jié)點(diǎn)值即可，代碼如下：

/**
 * 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 kthSmallest(TreeNode* root, int k) {
        int cnt = 0;
        stack<TreeNode*> sk;
        TreeNode* p = root;
        while (p || !sk.empty()) {
            while (p) {
                sk.push(p);
                p = p->left;
            }
            p = sk.top();
            sk.pop();
            ++cnt;
            if (cnt == k)
                return p->val;
            p = p->right;
        }
        return 0;
    }
};