【LeetCode】230. Kth Smallest Element in a BST 解題報告

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230. Kth Smallest Element in a BST / Medium

Given the root of a binary search tree, and an integer k, return the kth smallest value (1-indexed) of all the values of the nodes in the tree.

Example 1:

Input: root = [3,1,4,null,2], k = 1
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
Output: 3

Comstraints

  • The number of nodes in the tree is n.
  • 1 <= k <= n <= 10^4
  • 0 <= Node.val <= 10^4
  • Follow up: If the BST is modified often (i.e., we can do insert and delete operations) and you need to find the kth smallest frequently, how would you optimize?

Solution 1: Traverse and store

思路

因為 BST 如果用 inorder 去尋訪的話會是排序過的，所以我們先用 inorder 尋訪一次，並用陣列紀錄順序。接著直接取出 index k-1 的值就好。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    vector<int> nums;
    int kthSmallest(TreeNode* root, int k) {
        if(!root) return 0;
        traverse(root);
        return nums[k-1];
    }
    
    void traverse(TreeNode* root) {
        if(!root) return;
        
        traverse(root->left);
        nums.emplace_back(root->val);
        traverse(root->right);
    }
};

Solution 2: Iterate with Stack

思路

如果改用 Stack 來尋訪，可以在走到第 k 次時直接 reture 答案。

效能

Complexity

  • Time Complexity: O(N)
  • Space Complexity: O(N)

LeetCode Result

Code

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class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        stack<TreeNode*> node_stack;
        while(true) {
            while(root) {
                node_stack.push(root);
                root = root->left;
            }
            root = node_stack.top();
            node_stack.pop();
            if(--k == 0) return root->val;
            root = root->right;
        }
    }
};