LeetCode173 Binary Search Tree Iterator

最新推荐文章于 2023-02-25 23:52:55 发布

tigerhuli

最新推荐文章于 2023-02-25 23:52:55 发布

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分类专栏： LeetCode

本文链接：https://blog.youkuaiyun.com/m0_37518259/article/details/88351686

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本文介绍了一种在二叉搜索树（BST）上进行迭代的算法，通过两种方法实现了一个迭代器，能够平均在O(1)的时间复杂度下返回BST中的下一个最小数值，并使用O(h)的内存，其中h为树的高度。

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.

Example:

BSTIterator iterator = new BSTIterator(root);
iterator.next();    // return 3
iterator.next();    // return 7
iterator.hasNext(); // return true
iterator.next();    // return 9
iterator.hasNext(); // return true
iterator.next();    // return 15
iterator.hasNext(); // return true
iterator.next();    // return 20
iterator.hasNext(); // return false

Note:

next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.
You may assume that next() call will always be valid, that is, there will be at least a next smallest number in the BST when next() is called.

题源：here；完整实现：here

思路：

1 我们将二叉树全部存储到一个vector中；2 利用stack做到一边取数一边遍历二叉树。

解法1

class BSTIterator {
public:
	vector<int> nums;
	BSTIterator(TreeNode* root) {
		initN(root);
	}

	void initN(TreeNode *root) {
		if (!root) return;
		if (root->left) initN(root->left);
		nums.push_back(root->val);
		if (root->right) initN(root->right);
	}

	/** @return the next smallest number */
	int next() {
		int tmp = nums[0];
		nums.erase(nums.begin());
		return tmp;
	}

	/** @return whether we have a next smallest number */
	bool hasNext() {
		return nums.size();
	}
};

解法2

class BSTIterator2{
public:
	BSTIterator2(TreeNode* root) {
		addS(root);
	}

	/** @return the next smallest number */
	int next() {
		TreeNode *tmp = s.top();
		s.pop();
		if (tmp->right) addS(tmp->right);
		return tmp->val;
	}

	/** @return whether we have a next smallest number */
	bool hasNext() {
		return !s.empty();
	}
private:
	stack<TreeNode *> s;
	void addS(TreeNode *root) {
		while (root) {
			s.push(root);
			root = root->left;
		}
	}
};