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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

 

Constraints:

  • The number of nodes in the tree is n.
  • 1 <= k <= n <= 104
  • 0 <= Node.val <= 104

 

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?

Companies:
Amazon, Uber, Facebook

Related Topics:
Tree, Depth-First Search, Binary Search Tree, Binary Tree

Similar Questions:

Solution 1. In-order traversal (Recursive)

// OJ: https://leetcode.com/problems/kth-smallest-element-in-a-bst/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        function<int(TreeNode*)> inorder = [&](TreeNode *root) {
            if (!root) return -1;
            int val = inorder(root->left);
            if (val != -1) return val;
            if (--k == 0) return root->val;
            return inorder(root->right);
        };
        return inorder(root);
    }
};

Solution 1. In-order traversal (Iterative)

// OJ: https://leetcode.com/problems/kth-smallest-element-in-a-bst/
// Author: github.com/lzl124631x
// Time: O(N)
// Space: O(H)
class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        stack<TreeNode*> s;
        while (root || s.size()) {
            while (root) {
                s.push(root);
                root = root->left;
            }
            root = s.top();
            s.pop();
            if (--k == 0) return root->val;
            root = root->right;
        }
        return -1;
    }
};