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.
实现一个基于二叉搜索树的类iterator。初始化iterator为这棵树的根节点root.。调用next（）函数返回这棵树的下一个最小的值。
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class BSTIterator {
stack<TreeNode*> qu;
public:
BSTIterator(TreeNode *root) {
push_in_stack(root);
}
/** @return whether we have a next smallest number */
bool hasNext() {
return !qu.empty();
}
/** @return the next smallest number */
int next() {
TreeNode *tmpNode = qu.top();
qu.pop();
push_in_stack(tmpNode->right);
return tmpNode->val;
}
private:
void push_in_stack(TreeNode* node)
{
while(node)
{
qu.push(node);
node=node->left;
}
}
};
/**
* Your BSTIterator will be called like this:
* BSTIterator i = BSTIterator(root);
* while (i.hasNext()) cout << i.next();
*/while (i.hasNext()) cout << i.next();
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原文：http://blog.csdn.net/sinat_24520925/article/details/46793875