LeetCode 173. Binary Search Tree Iterator

原题链接在这里：https://leetcode.com/problems/binary-search-tree-iterator/

题目：

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.

题解：

用一个stack把所有最左边的node压进 stack中。

判断hasNext()时就是看stack 是否为空.

next()返回stack顶部top元素，若是top有右子树，就把右子树的所有最左node压入stack中.

constructor time complexity: O(h).

hashNext time complexity: O(1).

next time complexity: O(h).

Space: O(h), stack 最大为树的高度。

AC Java:

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

public class BSTIterator {
    Stack<TreeNode> stk;
    
    public BSTIterator(TreeNode root) {
        stk = new Stack<TreeNode>();
        while(root != null){
            stk.push(root);
            root = root.left;
        }
    }

    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        return !stk.isEmpty();
    }

    /** @return the next smallest number */
    public int next() {
        TreeNode top = stk.pop();
        TreeNode rightLeft = top.right;
        while(rightLeft != null){
            stk.push(rightLeft);
            rightLeft = rightLeft.left;
        }
        
        return top.val;
    }
}

/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */

类似Binary Tree Inorder Traversal, Inorder Successor in BST.