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.
題目題意感覺不好理解,查了下大概是按中序遍歷把節點非遞歸輸出
/** * Definition for binary tree * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */ class BSTIterator { public: stacks; BSTIterator(TreeNode *root) { while(root != NULL){ 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 * tmp = s.top(); s.pop(); if(tmp->right != NULL){ TreeNode* p = tmp->right; while(p){ s.push(p); p = p->left; } } return tmp->val; } }; /** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */