173. 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.

链接： http://leetcode.com/problems/binary-search-tree-iterator/

6/20/2017
7ms, 23%

有想法，但是代码主要是看别人的

思路是把inorder traversal拆开，运行next时候把右边的压进栈

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

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

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

    /** @return the next smallest number */
    public int next() {
        if (hasNext()) {
            TreeNode top = stack.pop();
            if (top.right != null) {
                inorder(top.right);
            }
            return top.val;
        }
        return -1;
    }
}

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

针对时间复杂度的解释：

The average time complexity of next() function is O(1) indeed in your case. As the next function can be called n times at most, and the number of right nodes in self.pushAll(tmpNode.right) function is maximal n in a tree which has n nodes, so the amortized time complexity is O(1).

https://discuss.leetcode.com/topic/6575/my-solutions-in-3-languages-with-stack

更多讨论

https://discuss.leetcode.com/category/181/binary-search-tree-iterator