学步园

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();
 */