173-二叉搜索树迭代器

Description

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.

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Note: next() and hasNext() should run in average O(1) time and uses O(h) memory, where h is the height of the tree.

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问题描述

实现二叉搜索树迭代器。迭代器初始化为树的根节点
调用next()返回BST中的下一个最小的树

注意:next()和hasNext()需要是O(1)的时间复杂度，O(h)的空间复杂度，h为树的高度。

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问题分析

使用stack保存二叉搜索树节点。

1. 构造函数中将从根节点到最左节点之间的节点入栈。
2. hasNext()判断栈是否空
3. next(),获取出栈节点，判断是否有右子树，若有，则将右子树根节点到其最左节点之间的节点入栈。返回出栈节点的值。

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解法

public class BSTIterator {

    private Stack<TreeNode> stack;

    public BSTIterator(TreeNode root) {
        stack = new Stack();
        TreeNode cur = root;
        while(cur != null){
            stack.push(cur);
            if(cur.left != null)    cur = cur.left;
            else    break;
        }
    }

    /** @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();
        TreeNode cur = node;

        // traversal right branch
        if(cur.right != null){
            cur = cur.right;
            while(cur != null){
                stack.push(cur);
                if(cur.left != null)    cur = cur.left;
                else    break;
            }
        }

        return node.val;
    }
}