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Leetcode[230] Kth Smallest Element in a BST

29 Nov 2015

###Task1 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).

###Python ####O(K)

# Definition for a binary tree node.
# class TreeNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution(object):
    def kthSmallest(self, root, k):
        """
        :type root: TreeNode
        :type k: int
        :rtype: int
        """
        stack = []
        prev = root
        count = 0
        while len(stack) or prev:
            while prev:
                stack.append(prev)
                prev = prev.left
            cur = stack.pop()
            count += 1
            if count == k:
                return cur.val
            prev = cur.right

###Java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {

	public void inorder(TreeNode root, ArrayList<Integer> list) {
		if (root == null) {
			return ;
		}
		inorder(root.left, list);
		list.add(root.val);
		inorder(root.right, list);

		return;
	}

    public int kthSmallest(TreeNode root, int k) {
        if (root == null) {
        	return 0;
        }

        ArrayList<Integer> list = new ArrayList<Integer>();
        inorder(root, list);

        return list.get(k - 1);
    }
}

###Points

如果BST节点TreeNode的属性可以扩展，则再添加一个属性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)

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