Leetcode - Binary Search Tree Iterator

《Leetcode - Binary Search Tree Iterator》

My code:

import java.util.ArrayList;

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

public class BSTIterator {
    private ArrayList<Integer> bst = null;
    private int index;
    public BSTIterator(TreeNode root) {
        bst = new ArrayList<Integer>();
        if (root == null) {
            index = Integer.MAX_VALUE;
            return;
        }
        this.index = 0;
        inOrder(root, bst);
    }
    
    private void inOrder(TreeNode root, ArrayList<Integer> bst) {
        if (root.left != null)
            inOrder(root.left, bst);
        bst.add(root.val);
        if (root.right != null)
            inOrder(root.right, bst);
    }
    /** @return whether we have a next smallest number */
    public boolean hasNext() {
        if (this.index < bst.size())
            return true;
        else
            return false;
    }

    /** @return the next smallest number */
    public int next() {
        return bst.get(index++);
    }
}

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

My test result:

《Leetcode - Binary Search Tree Iterator》 Paste_Image.png

现在刷题,做完之后,都忘记这道题目是干什么的了。。。是立刻就忘记。。。
这道题目,挺新颖的。我自己的做法,就是生成一个arraylist,按照inorder的顺序。
这些正好今天上课全部讲到了。
然后每次需要最小的,就取出index,并且往前进一格。
判断hasNext(), 就判断index是否 >= arrayList.size() 就行了。

然后有点麻烦的地方,就是如果root = null。
怎么办。
然后我学习到了,
原来构造函数中,也可以使用return

**
总结: inorder BST 出来的数列是从小到大排列的。 构造函数中可以使用 return
**

Anyway, Good luck, Richardo!

My code:

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

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

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

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

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

题目的要求是,
时间, 平均,O(1)
空间, O(h)

所以我的原做法,空间复杂度达到了O(n)

现在采用非递归的做法做。将左侧的都压到栈中。空间复杂度就达到了O(h)
然后,每次弹出一个,即为返回值。
同时需要将它的右子树的左侧一边的所有结点全部压到栈中。空间复杂度最恶劣的化可以达到O(n)
但是,从平均来看,访问n个元素,时间复杂度是O(n)
所以对于每一个next操作,复杂度平均下来,是O(1)

就是这个O(1) 卡住了我。。。
参考网页:
http://www.programcreek.com/2014/04/leetcode-binary-search-tree-iterator-java/

Anyway, Good luck, Richardo!

My code:

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

public class BSTIterator {
    Stack<TreeNode> st = new Stack<TreeNode>();
    public BSTIterator(TreeNode root) {
        TreeNode curr = root;
        while (curr != null) {
            st.push(curr);
            curr = curr.left;
        }
    }

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

    /** @return the next smallest number */
    public int next() {
        TreeNode ret = st.pop();
        TreeNode curr = ret;
        if (curr.right != null) {
            curr = curr.right;
            while (curr != null) {
                st.push(curr);
                curr = curr.left;   
            }
        }
        
        return ret.val;
    }
}

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

自己写了出来。主要就是 average O(1)

Anyway, Good luck, Richardo! — 09/06/2016

    原文作者:Richardo92
    原文地址: https://www.jianshu.com/p/d21af604282e#comments
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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