173. Binary Search Tree Iterator

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

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

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

Analysis:

Recall how iterative in-order traverse is done:

//sudo code
while (cur != null || stack is not empty)
    while(cur is not null)
        stack.push(cur);
        cur = cur.left;
    visit(stack.pop()); // visit node
    cur = cur.right;

We basically need to conduct a cycle starting at line "visit node" when "next()" is called till right before visit node line.

When "hasNext()" is called, we just see if stack is empty.

Complexity:

Time: O(1) on average, as in total we need to push N nodes in stack, while next is called n times, the complexity for each time is O(1)

Space: O(logn)

Code:

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

public class BSTIterator {

    Deque<TreeNode> stack;
    public BSTIterator(TreeNode root) {
        stack = new LinkedList<TreeNode>();
        while(root!=null)
        {
            stack.push(root);
            root = root.left;
        }
    }

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

    /** @return the next smallest number */
    public int next() {
        TreeNode returnNode = stack.pop();
        TreeNode cur = returnNode.right;
        while(cur != null)
        {
            stack.push(cur);
            cur = cur.left;
        }
        return returnNode.val;
    }
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */