Leetcode 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.

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?

Hint:

1. Try to utilize the property of a BST.
2. What if you could modify the BST node's structure?
3. The optimal runtime complexity is O(height of BST).

題目意思：

　　給定一棵二叉搜尋樹，找到第k小的元素

　　注意：

　　　　1、利用二叉搜尋樹的特性

　　　　2、修改二叉搜尋樹的節點結構

　　　　3、最優時間複雜度是0(樹的高度)

解題思路：

方法一：

　　二叉搜尋樹的特性：其中序遍歷是有序的。故中序遍歷訪問，訪問第k個元素即可。

                           

方法二：

　　利用分冶的方法。

• 統計根節點左子樹的節點個數cnt
• 如果cnt+1 = k，則根節點即為第k個最小元素
• 如果cnt+1 > k，則第k個最小元素在左子樹，root = root->left;
• 如果cnt+1 < k，則第k個最小元素在右子樹，root = root->right;
• 重複第一步

原始碼：

方法一：

class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        stack<TreeNode*> nodeStack;
        if(root == NULL) return -1;
        while(true){
            while(root){
                nodeStack.push(root);
                root = root->left;
            }
            TreeNode* node = nodeStack.top(); nodeStack.pop();
            if(k == 1) return node->val;
            else root = node->right,k--;
        }
    }
};

方法二：

class Solution {
public:
    int calcNodeSize(TreeNode* root){
        if( root == NULL) return 0;
        return 1 + calcNodeSize(root->left) + calcNodeSize(root->right);
    }
    
    int kthSmallest(TreeNode* root, int k) {
        if(root == NULL) return 0;
        int cnt = calcNodeSize(root->left);
        if(k == cnt + 1) return root->val;
        else if( k < cnt + 1 ) return kthSmallest(root->left,k);
        else return kthSmallest(root->right, k-cnt-1);
    }
};