[LeetCode 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.

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.

Solution:

Use the stack to store left most node in the BST.

Top element is the smalleset one, and pop it out. store its right child's left node.

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

public class BSTIterator {

    Stack<TreeNode> nodePath = new Stack<>();
    public BSTIterator(TreeNode root) {
        while(!nodePath.isEmpty())
            nodePath.clear();
        TreeNode n1 = root;
        while(n1!=null){
            nodePath.add(n1);
            n1 = n1.left;
        }
    }

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

    /** @return the next smallest number */
    public int next() {
        TreeNode n1 = nodePath.pop();
        int res = n1.val;
        n1 = n1.right;
        while(n1!=null){
            nodePath.push(n1);
            n1 = n1.left;
        }
        return res;
    }
}

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