Binary Search Tree Iterator


#1

Implement an iterator over a binary search tree (BST). Your iterator will be initialized with the root node of a BST.

Calling next() will return the next smallest number in the BST.

Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

Credits:
Special thanks to @ts for adding this problem and creating all test cases.


#2

The algorithms below uses a Stack to replicate an inorder traversal that gives an ordered list of elements in a binary search tree.

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */

type Stack []*TreeNode
		
func (s *Stack) Push(n *TreeNode) {
    *s = append(*s, n)
}

func (s *Stack) Pop() (n *TreeNode) {
    if s.Len()!=0{
         x := s.Len() - 1
         n = (*s)[x]
        *s = (*s)[:x]
        return n
    }
   
    return nil
}

func (s *Stack) Len() int {
    return len(*s)
}


type BSTIterator struct {
    stack Stack
}


func Constructor(root *TreeNode) BSTIterator {
    s := Stack{}
    b := BSTIterator{stack : s}
    b.pushAll(root)
    return b
}


/** @return the next smallest number */
func (this *BSTIterator) Next() int {    
    node := this.stack.Pop()
    this.pushAll(node.Right)
    return node.Val
}

func (this *BSTIterator) pushAll(node *TreeNode){
    for node!= nil{
        this.stack.Push(node)
        node = node.Left
    }
}


/** @return whether we have a next smallest number */
func (this *BSTIterator) HasNext() bool {
    return this.stack.Len()!=0
}


/**
 * Your BSTIterator object will be instantiated and called as such:
 * obj := Constructor(root);
 * param_1 := obj.Next();
 * param_2 := obj.HasNext();
 */

#3

Similar code that use a Stack, but in Java.

public class BSTIterator {
    Stack<TreeNode> stack;
    public BSTIterator(TreeNode root) {
       this.stack = new Stack<>();
       pushAll(root);
    }
    
    private void pushAll(TreeNode node){
        while(node!=null){
            stack.push(node);
            node = node.left;
        }
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty();
    }

    /** @return the next smallest number */
    public int next() {
       TreeNode cur = stack.pop();
       pushAll(cur.right);
       return cur.val;
    }
}