## 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.

You may assume k is always valid, 1 ≤ k ≤ BST's total elements.

### Analysis:

Remind the concept of BST (see here), a tree traversal will solve this problem.

The idea is to keep an array of length K, to hold the k smallest nodes. Each time when visiting a node, save it to the array if the array is not full. The end (or front depending on your save order) element is the Kth smallest node.

If insert and delete operations often happen, we can only check this array of length K, if any node in this array is deleted, search and insert the next element from the array's last node. If new node is inserted as the child of one node in the array, insert this node to its proper position of array, and move every other accordingly.

### Code(C++):

/** * 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: void search(TreeNode* root, int &k, vector<TreeNode*> &list_k){ if (root->left){ search(root->left, k, list_k);} k = k-1; if (k>=0){ list_k[k] = root; }else{ return; } if (root->right){search(root->right, k, list_k);} } int kthSmallest(TreeNode* root, int k) { vector<TreeNode*> list_k(k,NULL); search(root,k, list_k); return list_k[0]->val; } };

### Code(Python):

# Definition for a binary tree node. # class TreeNode(object): # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution(object): def search(self, root, k, list_k): if root.left: self.search(root.left, k, list_k) k[0] = k[0] -1 if k[0] >= 0: list_k.append(root) else: return if root.right: self.search(root.right, k, list_k) def kthSmallest(self, root, k): """ :type root: TreeNode :type k: int :rtype: int """ list_k = [] self.search(root, [k], list_k) return list_k[-1].val

How about maintain two bst, one keep the smallest k, one keep the rest n - k.

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