173. Binary Search Tree Iterator

题目：二叉搜索树（BST）迭代器

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.

题意：

实现一个二叉搜索树（BST）的迭代器。你的迭代器会使用BST的根节点初始化。

调用next()会返回BST中下一个最小的数字。

Note：

next()和hasNext()应该满足平均O(1)时间复杂度和O(h)空间复杂度，其中h是树的高度。

思路：

维护一个栈，从根节点开始，每次迭代地将根节点的左孩子压入栈，直到左孩子为空为止。调用next()方法时，弹出栈顶，如果被弹出的元素拥有右孩子，则以右孩子为根，将其左孩子迭代压栈。

代码：java版：8ms

/**
 * 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) {
        pushAll(root);
    }
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stack.isEmpty();
    }
    /** @return the next smallest number */
    public int next() {
        TreeNode tmpNode = stack.pop();
        pushAll(tmpNode.right);
        return tmpNode.val;
    }
    private void pushAll(TreeNode node) {
        for (; node!=null; stack.push(node), node = node.left);
    }
}
/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */

代码：C++版：26ms

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
    stack<TreeNode *> mystack;
public:
    BSTIterator(TreeNode *root) {
        pushAll(root);
    }
    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !mystack.empty();
    }
    /** @return the next smallest number */
    int next() {
        TreeNode *tmpNode = mystack.top();
        mystack.pop();
        pushAll(tmpNode->right);
        return tmpNode->val;
    }
private:
    void pushAll(TreeNode *node) {
        for (; node!=NULL; mystack.push(node), node = node->left);
    }
};
/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */

与以上C++版基本一致，pushAll循环实现细微差别：C++版：28ms

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class BSTIterator {
    stack<TreeNode *> mystack;
public:
    BSTIterator(TreeNode *root) {
        pushAll(root);
    }
    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !mystack.empty();
    }
    /** @return the next smallest number */
    int next() {
        TreeNode *tmpNode = mystack.top();
        mystack.pop();
        pushAll(tmpNode->right);
        return tmpNode->val;
    }
private:
    void pushAll(TreeNode *node) {
        while (node) {
            mystack.push(node);
            node = node->left;
        }
    }
};
/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */