173. Binary Search Tree Iterator

1.問題

• 實作BST的迭代器, constructor, next, hasNext

2.想法

• 用stack儲存BST的left主線上的node

  • last in first out的特性

• BST的left主線是最小值, 而left主線上node的每一個right node的left主線又是另一個list, 其大小介於該node的以及node的上一次中

  • 因此每次pop stack的元素時, 要再將top的right以及其left主線上的元素全部加入

3.程式碼

/**
 * 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) {
        TreeNode * curr = root;
        while (curr) {
            st.push(curr);
            curr = curr->left;
        }
    }

    /** @return whether we have a next smallest number */
    bool hasNext() {
        return !st.empty();
    }

    /** @return the next smallest number */
    int next() {
        int val = st.top()->val;
        TreeNode* node = st.top()->right;
        st.pop();
        while (node) {
            st.push(node);
            node = node->left;
        }
        return val;
    }
private:
    stack<TreeNode*> st;
};

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = BSTIterator(root);
 * while (i.hasNext()) cout << i.next();
 */

4.Performance