LintCode-Implement Iterator of Binary Search Tree
Design an iterator over a binary search tree with the following properties:
- Elements are visited in ascending order (i.e. an inorder traversal)
- next() and hasNext() queries run in O(1) time in average.
For the following binary search tree, inorder traversal by using iterator is [1, 6, 10, 11, 12]
10
/ \
1 11
\ \
6 12
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
* Example of iterate a tree:
* Solution iterator = new Solution(root);
* while (iterator.hasNext()) {
* TreeNode node = iterator.next();
* do something for node
* }
*/
public class Solution {
private Stack<TreeNode> nodeStack; //@param root: The root of binary tree.
public Solution(TreeNode root) {
nodeStack = new Stack<TreeNode>();
//Initialize first, then determine null, otherwise, the hasNext() function will cause problem.
if (root==null) return;
nodeStack.push(root);
TreeNode cur = root.left;
while (cur!=null){
nodeStack.push(cur);
cur = cur.left;
}
} //@return: True if there has next node, or false
public boolean hasNext() {
if (nodeStack.isEmpty()) return false;
else return true;
} //@return: return next node
public TreeNode next() {
if (nodeStack.isEmpty()) return null;
TreeNode next = nodeStack.pop();
if (next.right==null) return next;
else {
nodeStack.push(next.right);
TreeNode cur = next.right.left;
while (cur!=null){
nodeStack.push(cur);
cur = cur.left;
}
return next;
}
}
}
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