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230. Kth Smallest Element in a BST

Description

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?

Solutions

Solution 1

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def kthSmallest(self, root: Optional[TreeNode], k: int) -> int:
        stk = []
        while root or stk:
            if root:
                stk.append(root)
                root = root.left
            else:
                root = stk.pop()
                k -= 1
                if k == 0:
                    return root.val
                root = root.right

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Deque<TreeNode> stk = new ArrayDeque<>();
        while (root != null || !stk.isEmpty()) {
            if (root != null) {
                stk.push(root);
                root = root.left;
            } else {
                root = stk.pop();
                if (--k == 0) {
                    return root.val;
                }
                root = root.right;
            }
        }
        return 0;
    }
}

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    int kthSmallest(TreeNode* root, int k) {
        stack<TreeNode*> stk;
        while (root || !stk.empty()) {
            if (root) {
                stk.push(root);
                root = root->left;
            } else {
                root = stk.top();
                stk.pop();
                if (--k == 0) return root->val;
                root = root->right;
            }
        }
        return 0;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func kthSmallest(root *TreeNode, k int) int {
    stk := []*TreeNode{}
    for root != nil || len(stk) > 0 {
        if root != nil {
            stk = append(stk, root)
            root = root.Left
        } else {
            root = stk[len(stk)-1]
            stk = stk[:len(stk)-1]
            k--
            if k == 0 {
                return root.Val
            }
            root = root.Right
        }
    }
    return 0
}

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