? leetcode 173. Binary Search Tree Iterator 设计迭代器（搜索树）--------- java

Posted 2020-08-22 xiaoba1203

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.

 

设计一个二叉搜索树的迭代器。要求其中的next()与hasNext()是平均O(1)的时间复杂度，O(h)的空间复杂度，h是树高。

 

1、用栈来实现，栈中存储的是当前路径的左孩子。

 

/**
 * 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();
        if (root == null){
            return ;
        }
        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() {
        TreeNode node = stack.pop();
        int ans = node.val;
        if (node.right != null){
            node = node.right;
            while (node != null){
                stack.push(node);
                node = node.left;
            }
        }
        return ans;
    }
    
    
}

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

 

 

2、用list实现。直接排序然后存储在list中，代码简单高效。（参考discuss）。

这种方法虽然比上面的方法快并且简单，但是使用的空间是O（N)的空间，比上一个多，如果上一个题意中说明该设计类只能用O(h)的空间，那么这种解法就不对了。

ArrayDeque<Integer> list;

public BSTIterator(TreeNode root) {
    list = new ArrayDeque<Integer>();
    inorderTraverse(root);
}

void inorderTraverse(TreeNode root)
{
    if(root == null)
        return;
    inorderTraverse(root.left);
    list.addLast(root.val);
    inorderTraverse(root.right);
}

/** @return whether we have a next smallest number */
public boolean hasNext() {
    if(list.isEmpty())
        return false;
    else
        return true;
}

/** @return the next smallest number */
public int next() {
    return list.removeFirst();
}