Kth smallest Element in a BST

Given A binary search tree, write a function to kthSmallest find the k-th smallest element in it.

Note:

You may assume k are always valid, 1≤k≤bst's total elements.

Problem-solving idea: The number of K small in BST (binary sort tree) is obtained. Method One: Traverse the whole binary tree, then sort, output the Val value of the K-node. The code is as follows (where we take the stack, advanced back out, right to left.) First-order traversal of the entire binary tree):

Class Solution {public:    int kthsmallest (treenode* root, int k) {        if (root==null) return 0;        Stack<treenode*> Qu;        vector<int> Res;        Qu.push (root);        while (!qu.empty ())        {            treenode* node=qu.top ();            Qu.pop ();            Res.push_back (node->val);             if (node->right!=null)            {                 qu.push (node->right);            }             if (node->left!=null)            {                 qu.push (node->left);            }        }        Sort (Res.begin (), Res.end ());        return res[k-1];}    ;

Method Two:

In a binary search tree, find the K element.

The algorithm is as follows:

1. Count left subtree elements.

2, left+1 = k, the root node is the K element

Left >=k, the first k element is

Left +1 <k, it is converted to the right subtree, looking for the k-left-1 element.

The code is as follows:

Class Solution {public:    int treeSize (treenode* root)    {       if (root==null) return 0;       Return 1+treesize (Root->left) +treesize (root->right);    }    int kthsmallest (treenode* root, int k) {        if (root==null) return 0;        int leftsize=treesize (root->left);        if (k==leftsize+1)        return root->val;        else if (leftsize>=k)        return Kthsmallest (Root->left, k);        else         return kthsmallest (Root->right, k-leftsize-1);            };

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Kth smallest Element in a BST