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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

  • k == root.leftCnt+1, 则返回 root
  • 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;
}
}