Design an iterator over a binary search tree with the following rules:
- Elements are visited in ascending order (i.e. an in-order traversal)
- next() and hasNext() queries run in O(1) time in average.
Example
For the following binary search tree, in-order traversal by using iterator is [1, 6, 10, 11, 12]
10
/ \
1 11
\ \
6 12
Challenge
Extra memory usage O(h), h is the height of the tree.
Super Star: Extra memory usage O(1)
仍然考的是中序遍历，但是是非递归实现。其实这道题等价于写一个二叉树中序遍历的迭代器。需要内置一个栈，一开始先存储到最左叶子节点的路径。在遍历的过程中，只要当前节点存在右子树，则进入右子树，存储从此处开始到当前子树里最左叶子节点的路径。
/**
* 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> stack = new Stack<>();
private TreeNode curt;
// @param root: The root of binary tree.
public Solution(TreeNode root) {
curt = root;
}
//@return: True if there has next node, or false
public boolean hasNext() {
return (curt != null || !stack.isEmpty()); //important to judge curt != null
}
//@return: return next node
public TreeNode next() {
while (curt != null) {
stack.push(curt);
curt = curt.left;
}
curt = stack.pop();
TreeNode node = curt;
curt = curt.right;
return node;
}
}
hasNext() 函数中的判断语句，不能漏掉 curt != null。