[LeetCode 230] Kth Smallest Element in a BST

标签：leetcode   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).

solution:

Inorder traverse, get kth element from that result. complexity is O(n)

public class Solution {
    List<Integer> path = new ArrayList<>();
    public int kthSmallest(TreeNode root, int k) {
        inorder(root);
        return path.get(k-1);
    }
    public void inorder(TreeNode root) {
        if(root != null){
            inorder(root.left);
            path.add(root.val);
            inorder(root.right);
        }
    }
}

[LeetCode 230] Kth Smallest Element in a BST

标签：leetcode   bst   

原文地址：http://blog.csdn.net/sbitswc/article/details/48544731