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

  1. // Kth Smallest Element in a BST
  2. // Time Complexity: O(k), Space Complexity: O(h)
  3. public class Solution {
  4.     public int kthSmallest(TreeNode root, int k) {
  5.         Stack s = new Stack<>();
  6.         TreeNode p = root;
  8.         while (!s.empty() || p != null) {
  9.             if (p != null) {
  10.                 s.push(p);
  11.                 p = p.left;
  12.             } else {
  13.                 p = s.pop();
  14.                 --k;
  15.                 if (k == 0) {
  16.                     return p.val;
  17.                 }
  18.                 p = p.right;
  19.             }
  20.         }
  21.         return -1;
  22.     }
  23. }

原文: https://soulmachine.gitbooks.io/algorithm-essentials/content/java/binary-tree/bst/kth-smallest-element-in-a-bst.html