[LeetCode] 173.Binary Search Tree Iterator 二叉搜索树迭代器
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.
Credits:
Special thanks to @ts for adding this problem and creating all test cases.
这道题主要就是考二叉树的中序遍历的非递归形式，需要额外定义一个栈来辅助，二叉搜索树的建树规则就是左<根<右，用中序遍历即可从小到大取出所有节点。代码如下：
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/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class BSTIterator {public: BSTIterator(TreeNode *root) { while (root) { s.push(root); root = root->left; } } /** @return whether we have a next smallest number */ bool hasNext() { return !s.empty(); } /** @return the next smallest number */ int next() { TreeNode *n = s.top(); s.pop(); int res = n->val; if (n->right) { n = n->right; while (n) { s.push(n); n = n->left; } } return res; }private: stack<TreeNode*> s;};/** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */ |
原文链接：https://www.cnblogs.com/grandyang/p/4231455.html
