[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(); */