binary-tree/bst/kth-smallest-element-in-a-bst

Kth Smallest Element in a BST

描述

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:

• Try to utilize the property of a BST.
• What if you could modify the BST node's structure?
• The optimal runtime complexity is O(height of BST).

分析

最简单的办法，中序遍历，即可以得到递增序列，从而可以找到第 k 大的元素。时间复杂度O(k)。

如果能够修改节点的数据结构TreeNode，可以增加一个字段leftCnt，表示左子树的节点个数。设当前节点为root，

• 若 k == root.leftCnt+1, 则返回 root
• 若 k > node.leftCnt, 则 k -= root.leftCnt+1, root=root.right
• 否则，node = node.left

算法复杂度为O(height of BST)。

解法 1

// Kth Smallest Element in a BST
// Time Complexity: O(k), Space Complexity: O(h)
public class Solution {
    public int kthSmallest(TreeNode root, int k) {
        Stack<TreeNode> s = new Stack<>();
        TreeNode p = root;

        while (!s.empty() || p != null) {
            if (p != null) {
                s.push(p);
                p = p.left;
            } else {
                p = s.pop();
                --k;
                if (k == 0) {
                    return p.val;
                }
                p = p.right;
            }
        }
        return -1;
    }
}