#leetcode#Kth Smallest Element in a BST

标签：leetcode

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?

思路， in order traversal，

注意第一个元素的判断， 设root为当前节点， 则 root.left == null && count == 0，就保证了root为 1st smallest element，例子：

   4

  /

/

     2

  /

/

     1

    4

  /

/

       2

         \

           \

    3

上面两个bst中第一个节点分别为 1 和 2

/**
 * 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 kthSmallest(TreeNode root, int k) {
        List<Integer> count = new ArrayList<Integer>();
        count.add(0);
        helper(root, count, k);
        return count.get(1);
    }
    
    private void helper(TreeNode root, List<Integer> count, int target){
        if(root == null){
            return;
        }
        
        helper(root.left, count, target);
        
        if(root.left == null && count.get(0) == 0){
            count.set(0, 1);
        }else{
            count.set(0, count.get(0) + 1);
        }
        if(count.get(0) == target){
            count.add(root.val);
            return;
        }
        
        helper(root.right, count, target);
    }
}

#leetcode#Kth Smallest Element in a BST

标签：leetcode

原文地址：http://blog.csdn.net/chibaoneliuliuni/article/details/46721285