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).
思路分析:基本就是实现BST的中序遍历。BST中序遍历得到有序的数组,能够easy知道第k小数。下面给出了递归实现,借助counter记录当前遍历过的node数目。当counter==k时就能够返回。
当然也能够借助栈实现迭代的解法。
Follow up: 进一步优化。我们能够在节点中额外保留一些信息: 左子树的大小. 在插入删除时也同一时候维护左子树的大小.进行查找时能够比較左子树size和k的大小,就能够知道要找的node在左边还是右边。通过分治法的思路加速. 时间复杂度为O(h)。
下面的伪代码摘录自 http://bookshadow.com/weblog/2015/07/02/leetcode-kth-smallest-element-bst/
假设BST节点TreeNode的属性能够扩展,则再加入一个属性leftCnt,记录左子树的节点个数
记当前节点为node
当node不为空时循环:
若k == node.leftCnt + 1:则返回node
否则。若k > node.leftCnt:则令k -= node.leftCnt + 1,令node = node.right
否则,node = node.left
上述算法时间复杂度为O(BST的高度)
当node不为空时循环:
若k == node.leftCnt + 1:则返回node
否则。若k > node.leftCnt:则令k -= node.leftCnt + 1,令node = node.right
否则,node = node.left
上述算法时间复杂度为O(BST的高度)
AC 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 res = 0; public int counter = 0; public int kthSmallest(TreeNode root, int k) { inOrderK(root, k); return res; } public void inOrderK(TreeNode root, int k){ if(root.left!=null) inOrderK(root.left, k); counter++; if(counter == k) { res = root.val; return; } if(root.right!=null) inOrderK(root.right, k); } }