Tree - 230. Kth Smallest Element in a BST

时间:2022-07-25
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230. Kth Smallest Element in a BST

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.

Example 1:

Input: root = [3,1,4,null,2], k = 1
   3
  / 
 1   4
  
   2
Output: 1

Example 2:

Input: root = [5,3,6,2,4,null,null,1], k = 3
       5
      / 
     3   6
    / 
   2   4
  /
 1
Output: 3

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:

由于bst的性质,所以bst的中序遍历,就是把bst从小到大输出,这样就能很容易找到第k小的数。

代码:

go:

/**

 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func kthSmallest(root *TreeNode, k int) int {
    stack := list.New()  
    cur := root
    
    for stack.Len() != 0 || cur != nil{
        if cur != nil {
            stack.PushBack(cur)
            cur = cur.Left
        } else {
            // visit
            e := stack.Back()
            stack.Remove(e)
            cur = e.Value.(*TreeNode)
            k--
            if 0 == k {
                return cur.Val
            }
            cur = cur.Right
        } 
    }
    
    return -1  // corner case
}

/*func kthSmallest(root *TreeNode, k int) int {
    stack := []*TreeNode{}
    t := root
	for t != nil || len(stack) != 0 {
		for t != nil {
			stack = append(stack, t)
			t = t.Left
		}
		t, stack = stack[len(stack)-1], stack[:len(stack)-1]
		k--
		if k == 0 {
			return t.Val
		}
		t = t.Right
	}
	return -1
}*/