leetcode 230: Kth Smallest Element in a BST

Kth Smallest Element in a BST

Total Accepted: 3655 Total Submissions: 12149

Given a binary search tree, write a function kthSmallest to find thekth 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?

[思路]

能够用递归和非递归. 这里我用了非递归的方法, 类似于 binary tree iterator. 顺便提一下, 我看了下网上的递归方法,时间复杂度非常高,每次都要算tree size. 感觉即使是递归,也应该还能optimize.

Follow up 挺有意思, 须要在节点中保留一些额外的信息: 左子树的大小. 在插入删除时也要同一时候维护左子树的大小. 再查找时,能够用二分. 时间复杂度为O(h)

[CODE]

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