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.
中序遍历的思路。
用栈来保存从根到最左侧叶子节点的路径,栈最上面的结点是最小的结点,每次取next,都是取栈最上面的结点,然后把剩余结点到最左侧叶子节点的路径放入栈中。
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class BSTIterator { public: stack<TreeNode*> stk; BSTIterator(TreeNode *root) { while(!stk.empty()) stk.pop(); while(root){ stk.push(root); root = root->left; } } /** @return whether we have a next smallest number */ bool hasNext() { return !stk.empty(); } /** @return the next smallest number */ int next() { TreeNode* tmp = stk.top(); int res = tmp->val; stk.pop(); tmp = tmp->right; while(tmp) { stk.push(tmp); tmp = tmp->left; } return res; } }; /** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */
Leetcode: Binary Search Tree Iterator
原文地址:http://blog.csdn.net/doc_sgl/article/details/42372251