LeetCode - Medium - 173. Binary Search Tree Iterator

超、凢脫俗 2022-10-16 15:20 160阅读 0赞

## Topic ##

*  Stack
 *  Tree
 *  Design

## Description ##

[https://leetcode.com/problems/binary-search-tree-iterator/][https_leetcode.com_problems_binary-search-tree-iterator]

Implement the `BSTIterator` class that represents an iterator over the **in-order traversal** of a binary search tree (BST):

*  `BSTIterator(TreeNode root)` Initializes an object of the `BSTIterator` class. The `root` of the BST is given as part of the constructor. The pointer should be initialized to a non-existent number smaller than any element in the BST.
 *  `boolean hasNext()` Returns `true` if there exists a number in the traversal to the right of the pointer, otherwise returns `false`.
 *  `int next()` Moves the pointer to the right, then returns the number at the pointer.

Notice that by initializing the pointer to a non-existent smallest number, the first call to `next()` will return the smallest element in the BST.

You may assume that `next()` calls will always be valid. That is, there will be at least a next number in the in-order traversal when `next()` is called.

**Example 1:**

![743f55b78d358ff4ba5fbb7d8088a0d9.png][]

Input
    ["BSTIterator", "next", "next", "hasNext", "next", "hasNext", "next", "hasNext", "next", "hasNext"]
    [[[7, 3, 15, null, null, 9, 20]], [], [], [], [], [], [], [], [], []]
    Output
    [null, 3, 7, true, 9, true, 15, true, 20, false]
    
    Explanation
    BSTIterator bSTIterator = new BSTIterator([7, 3, 15, null, null, 9, 20]);
    bSTIterator.next();    // return 3
    bSTIterator.next();    // return 7
    bSTIterator.hasNext(); // return True
    bSTIterator.next();    // return 9
    bSTIterator.hasNext(); // return True
    bSTIterator.next();    // return 15
    bSTIterator.hasNext(); // return True
    bSTIterator.next();    // return 20
    bSTIterator.hasNext(); // return False

**Constraints:**

*  The number of nodes in the tree is in the range  \[ 1 , 1 0 5 \] \[1, 10^5\] \[1,105\].
 *   0 < = N o d e . v a l < = 1 0 6 0 <= Node.val <= 10^6 0<=Node.val<=106
 *  At most  1 0 5 10^5 105 calls will be made to `hasNext`, and `next`.

**Follow up:**

*  Could you implement `next()` and `hasNext()` to run in average O(1) time and use O(h) memory, where h is the height of the tree?

## Analysis ##

主要考察BST的中序遍历的非递归法。

方法一：我写的

方法二：别人写的

方法三：别人写的2

## Submission ##

import java.util.LinkedList;
    
    import com.lun.util.BinaryTree.TreeNode;
    
    public class BinarySearchTreeIterator { 
    
    	//方法一：我写的
    	public static class BSTIterator { 
    
    		private LinkedList<Object[]> stack;
    		
    	    public BSTIterator(TreeNode root) { 
    	    	stack = new LinkedList<>();
    	    	stack.push(new Object[] { root, 0});
    	    }
    	    
    	    public int next() { 
    	    	while(!stack.isEmpty()) { 
    	    		Object[] arr = stack.peek();
    	    		TreeNode node = (TreeNode)arr[0];
    	    		int state = (int)arr[1];
    	    		
    	    		if(state == 0) { 
    	    			if(node.left != null)
    	    				stack.push(new Object[] { node.left, 0});
    	    			arr[1] = 1;
    	    		}else{ 
    	    			stack.pop();
    	    			if(node.right != null) 
    	    				stack.push(new Object[] { node.right, 0});
    	    			return node.val;
    	    		}
    	    	}
    	    	return -1;
    	    }
    	    
    	    public boolean hasNext() { 
    	    	return !stack.isEmpty();
    	    }
    	}
    	
    	//方法二：别人写的
    	public static class BSTIterator2 { 
    
    		private LinkedList<TreeNode> stack;
    		private TreeNode pointer;
    		
    	    public BSTIterator2(TreeNode root) { 
    	    	stack = new LinkedList<>();
    	    	pointer = root;
    	    }
    	    
    	    public int next() { 
    	    	while(hasNext()) { 
    	    		if(pointer != null) { 
    	    			stack.push(pointer);
    	    			pointer = pointer.left;
    	    		}else{ 
    	    			TreeNode node = stack.pop();
    	    			pointer = node.right;
    	    			return node.val;
    	    		}
    	    	}
    	    	return -1;//或抛异常
    	    }
    	    
    	    public boolean hasNext() { 
    	    	return !stack.isEmpty() || pointer != null;
    	    }
    	}
    	
    	//方法三：别人写的2
    	public static class BSTIterator3 { 
    
    	    private LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
    	    
    	    public BSTIterator3(TreeNode root) { 
    	        pushAll(root);
    	    }
    
    	    /** @return whether we have a next smallest number */
    	    public boolean hasNext() { 
    	        return !stack.isEmpty();
    	    }
    
    	    /** @return the next smallest number */
    	    public int next() { 
    	        TreeNode tmpNode = stack.pop();
    	        pushAll(tmpNode.right);
    	        return tmpNode.val;
    	    }
    	    
    	    private void pushAll(TreeNode node) { 
    	        for (; node != null; stack.push(node), node = node.left);
    	    }
    	}
    	
    	
    }

## Test ##

import static org.junit.Assert.*;
    import org.junit.Test;
    
    import com.lun.medium.BinarySearchTreeIterator.BSTIterator;
    import com.lun.medium.BinarySearchTreeIterator.BSTIterator2;
    import com.lun.medium.BinarySearchTreeIterator.BSTIterator3;
    import com.lun.util.BinaryTree;
    import com.lun.util.BinaryTree.TreeNode;
    
    public class BinarySearchTreeIteratorTest { 
    
    	@Test
    	public void test() { 
    		TreeNode root = BinaryTree.integers2BinaryTree(7, 3, 15, null, null, 9, 20);
    		
    		BSTIterator bSTIterator = new BSTIterator(root);
    
    		assertEquals(3, bSTIterator.next());
    		assertEquals(7, bSTIterator.next());
    		assertTrue(bSTIterator.hasNext());
    		assertEquals(9, bSTIterator.next());
    		assertTrue(bSTIterator.hasNext());
    		assertEquals(15, bSTIterator.next());
    		assertTrue(bSTIterator.hasNext());
    		assertEquals(20, bSTIterator.next());
    		assertFalse(bSTIterator.hasNext());
    	}
    	
    	@Test
    	public void test2() { 
    		TreeNode root = BinaryTree.integers2BinaryTree(7, 3, 15, null, null, 9, 20);
    		
    		BSTIterator2 bSTIterator = new BSTIterator2(root);
    		
    		assertEquals(3, bSTIterator.next());
    		assertEquals(7, bSTIterator.next());
    		assertTrue(bSTIterator.hasNext());
    		assertEquals(9, bSTIterator.next());
    		assertTrue(bSTIterator.hasNext());
    		assertEquals(15, bSTIterator.next());
    		assertTrue(bSTIterator.hasNext());
    		assertEquals(20, bSTIterator.next());
    		assertFalse(bSTIterator.hasNext());
    	}
    	
    	@Test
    	public void test3() { 
    		TreeNode root = BinaryTree.integers2BinaryTree(7, 3, 15, null, null, 9, 20);
    		
    		BSTIterator3 bSTIterator = new BSTIterator3(root);
    		
    		assertEquals(3, bSTIterator.next());
    		assertEquals(7, bSTIterator.next());
    		assertTrue(bSTIterator.hasNext());
    		assertEquals(9, bSTIterator.next());
    		assertTrue(bSTIterator.hasNext());
    		assertEquals(15, bSTIterator.next());
    		assertTrue(bSTIterator.hasNext());
    		assertEquals(20, bSTIterator.next());
    		assertFalse(bSTIterator.hasNext());
    	}
    }

[https_leetcode.com_problems_binary-search-tree-iterator]: https://leetcode.com/problems/binary-search-tree-iterator/
[743f55b78d358ff4ba5fbb7d8088a0d9.png]: /images/20221014/f1f8e4b124964bfda2fc1d67a8e21b5a.png