LeetCode173-Binary Search Tree Iterator

LeetCode 173 : Binary Search Tree Iterator

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.

來源： https://leetcode.com/problems/binary-search-tree-iterator/

問題描述：

實現一個二叉搜索樹（BST）的迭代器。你的迭代器會使用BST的根節點初始化。

調用next()會返回BST中下一個最小的數字。

注意：next()和hasNext()應該滿足平均O(1)時間複雜度和O(h)空間複雜度，其中h是樹的高度。

解題思路：

創建一個棧，從根節點開始，每次迭代地將根節點的左孩子壓入棧，直到左孩子爲空爲止。

調用next()方法時，彈出棧頂，如果被彈出的元素擁有右孩子，則以右孩子爲根，將其左孩子迭代壓棧。

這裏從 http://yuanhsh.iteye.com/blog/2173429 這篇文章這裏借了兩張圖，很好的講述瞭解題思路。

源代碼：

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

    private Stack<TreeNode> nodeStack;
    public BSTIterator(TreeNode root) {
        nodeStack = new Stack<TreeNode>();
        while(null != root){
            nodeStack.push(root);
            root = root.left;
        }
    }
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
       return !nodeStack.isEmpty();
    }
    /** @return the next smallest number */
    public int next() {
        TreeNode node = nodeStack.pop();
        int result = node.val;
        if(null != node.right){
            node = node.right;
            while(null != node){
                nodeStack.push(node);
                node = node.left;
            }
        }
        return result;
    }
}
/**
 * Your BSTIterator will be called like this:
 * BSTIterator i = new BSTIterator(root);
 * while (i.hasNext()) v[f()] = i.next();
 */