leetcode173- Binary Search Tree Iterator- medium

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.

把迭代式中序遍历拆解在几个过程里。用Stack结构存储中间信息。迭代式中序遍历就是一开始一直把左边结点一路推到底，然后开始回吐到中间结点的时候，再开始对右节点做同样的左推到底的处理。写在这个API里的话，就是：

1.初始化：开栈，左推到底。

2.判断有无：看栈空不

3.拿下一个：pop，顺便每pop出一个中间节点，都要开始对它的右结点再做左推到底的事情了。

开销：因为分摊了，每次的时间复杂度的确是平均O(1)的，一共推拿过n个结点，你调用也是调用n次。

实现：

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

public class BSTIterator {
    
    private Stack<TreeNode> stack = new Stack<TreeNode>();
    
    public BSTIterator(TreeNode root) {
        pushLeftDown(root);
    }

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

    /** @return the next smallest number */
    public int next() {
        TreeNode node = stack.pop();
        pushLeftDown(node.right);
        return node.val;
    }
    
    private void pushLeftDown(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();
 */

2.散的啰嗦版

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

public class BSTIterator {
    
    private Stack<TreeNode> stack;
    
    public BSTIterator(TreeNode root) {
        stack = new Stack<TreeNode>();
        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 (stack.isEmpty()) {
            return -1;
        }
        TreeNode node = stack.pop();
        TreeNode right = node.right;
        while (right != null) {
            stack.push(right);
            right = right.left;
        }
        return node.val;
    }
}

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