Binary Search Tree Iterator

LeetCode 173–Binary Search Tree Iterator

Problem

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.

题目大意

实现一个用二叉排序树根节点初始化的二叉排序树迭代器。

调用 next() 方法返回二叉排序树中最小的下一个数。

注意：next() 和 hasNext()方法的平均时间复杂度为O(1) ，空间复杂度为 O(h)，h代表树的高度。

思路

该题目考察二叉树的中序遍历，步骤如下：
1. 初始化时，根节点进栈，左子树进栈，初始化结束。
2. 调用 hasNext()方法时，判断栈是否为空，若为空，则返回false，否则，返回true。
3. 调用next()方法时，栈顶元素出栈。并判断出栈元素是否有又子树，若存在，则右子树进栈，右子树的左子树进栈。返回出栈元素的值。

代码实现–Java

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

public class BSTIterator {

private Stack<TreeNode> stack = new Stack<TreeNode>();
public BSTIterator(TreeNode root) {
while(root != null){
stack.push(root);
root = root.left;
}
}

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

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

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

结果