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/

题解：

二叉搜索树iterator，要求O(1)的next()和hasNext()。可以用in-order traversal。

再仔细看一看，要求O(h)的memory，这样二刷的时候还要再仔细想一想。还有Morris Traversal要学习.

Time Complexity of next() and hasNext() - O(1)， Space Complexity - O(n)

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

public class BSTIterator {
    private Queue<Integer> queue;                                  //left, mid ,right

    public BSTIterator(TreeNode root) {
        this.queue = new LinkedList<>();
        inOrderTraversal(root);
    }
    
    private void inOrderTraversal(TreeNode root) {
        if(root == null)
            return;
            
        Stack<TreeNode> stack = new Stack<>();
        while(root != null || !stack.isEmpty()) {
            if(root != null) {
                stack.push(root);
                root = root.left;
            } else {
                root = stack.pop();
                queue.offer(root.val);
                root = root.right;
            }
        }
    }

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

    /** @return the next smallest number */
    public int next() {
        if(hasNext())
            return queue.poll();
        return Integer.MAX_VALUE;    
    }
}

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

 

二刷:

一刷写得不智慧。注意这里题目要求是amortized complexity - O(1)。是total expense of one operation。所以我们可以用把in-order traversal分解为左右两部分，然后分别放入stack中，就可以满足题目要求了。

为什么是amortized O(1)呢？ 因为在我们遍历整个树的过程中，对每个节点都只push 1次，所以对于遍历n个节点的树，我们总的expense是 O(n)，那amortized complexity就等于 O(1)了。

空间复杂度的减少，最好还是用Morris-traversal，下一次一定要好好写一遍。

Java:

Time Complexity: next() - amortized O(1),   hasNext() - O(1)， Space Complexity - O(h)

/**
 * 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<>();
        inorder(root);
    }

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

    /** @return the next smallest number */
    public int next() {
        TreeNode root = stack.pop();
        inorder(root.right);
        return root.val;
    }
    
    private void inorder(TreeNode root) {
        while (root != null) {
            stack.push(root);
            root = root.left;
        }
    }
}

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

 

Reference：

https://leetcode.com/discuss/20001/my-solutions-in-3-languages-with-stack

https://leetcode.com/discuss/20101/ideal-solution-using-stack-java

https://leetcode.com/discuss/30207/my-simple-solution-here

https://leetcode.com/discuss/23721/morris-traverse-solution

http://stackoverflow.com/questions/15079327/amortized-complexity-in-laymans-terms

https://en.wikipedia.org/wiki/Amortized_analysis