【LeetCode】230. Kth Smallest Element in a BST (2 solutions)

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?

Show Hint 

Credits:
Special thanks to @ts for adding this problem and creating all test cases.

解法一：递归中序遍历，必须全部遍历

/**
 * 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) {
        vector<int> ret;
        inOrder(root, ret);
        return ret[k-1];
    }
    void inOrder(TreeNode* root, vector<int>& ret)
    {
        if(root)
        {
            inOrder(root->left, ret);
            ret.push_back(root->val);
            inOrder(root->right, ret);
        }
    }
};

解法二：迭代中序遍历，遍历到第k个元素停止

/**
 * 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) {
        vector<int> ret;
        stack<TreeNode*> stk;
        stk.push(root);
        TreeNode* cur = root;
        while(cur->left)
        {
            stk.push(cur->left);
            cur = cur->left;
        }
        while(!stk.empty())
        {
            TreeNode* top = stk.top();
            stk.pop();
            ret.push_back(top->val);
            if(ret.size() == k)
                break;
            if(top->right)
            {
                TreeNode* cur = top->right;
                stk.push(cur);
                while(cur->left)
                {
                    stk.push(cur->left);
                    cur = cur->left;
                }
            }
        }
        return ret[k-1];
    }
};