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.

Credits:

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

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

public class BSTIterator {
private TreeNode node = null;
private Stack<TreeNode> stack = new Stack<>();

public BSTIterator(TreeNode root) {
node=root;
}

/** @return whether we have a next smallest number */
public boolean hasNext() {
return !(node==null&&stack.empty());
}

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

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