标签:ret uri track == return treenode bsp false cal
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.
/** * 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) { i=0; if(NULL==root) { length=0; } else { stack<TreeNode*> nodes; TreeNode* cur=root; while(!nodes.empty() || cur) { while(cur) { nodes.push(cur); cur=cur->left; } cur=nodes.top(); nodesValue.push_back(cur->val); nodes.pop(); cur=cur->right; } length=nodesValue.size(); } } /** @return whether we have a next smallest number */ bool hasNext() { if(i<length) return true; return false; } /** @return the next smallest number */ int next() { return nodesValue[i++]; } private: vector<int> nodesValue; int i; int length; }; /** * Your BSTIterator will be called like this: * BSTIterator i = BSTIterator(root); * while (i.hasNext()) cout << i.next(); */
leetCode(24):Binary Search Tree Iterator
标签:ret uri track == return treenode bsp false cal
原文地址:http://www.cnblogs.com/cynchanpin/p/7084792.html