[LeetCode 解题报告]230. 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:

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?

Method 1.

/**
 * 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 = count(root->left);
        if (k <= cnt)
            return kthSmallest(root->left, k);
        else if (k > cnt+1)
            return kthSmallest(root->right, k-cnt-1);
        else
            return root->val;
    }
    
    int count (TreeNode* root) {
        if (root == NULL)
            return 0;
        return count(root->left) + count(root->right) + 1;
    }
};
    

Method 2.

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