20150708 lintcode 总结 Binary Search Tree Iterator

最新推荐文章于 2019-08-29 23:09:36 发布

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最新推荐文章于 2019-08-29 23:09:36 发布

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分类专栏： lintcode 文章标签： lintcode

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本文介绍如何设计一个在平均情况下实现O(1)时间复杂度的二叉搜索树迭代器，支持in-order遍历并能高效处理树的高度。

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Binary Search Tree Iterator

Design an iterator over a binary search tree with the following rules:

Elements are visited in ascending order (i.e. an in-order traversal)
next() and hasNext() queries run in O(1) time in average.Example

For the following binary search tree, in-order traversal by using iterator is [1, 6, 10, 11, 12]

Challenge

Extra memory usage O(h), h is the height of the tree.

Super Star: Extra memory usage O(1)

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 * Example of iterate a tree:
 * Solution iterator = new Solution(root);
 * while (iterator.hasNext()) {
 *    TreeNode node = iterator.next();
 *    do something for node
 * } 
 */
public class Solution {
	public Stack myStack = new Stack();
	public TreeNode current;
    //@param root: The root of binary tree.
    public Solution(TreeNode root) {
        // write your code here
    	current = root;
    	return;
    }

    //@return: True if there has next node, or false
    public boolean hasNext() {
        // write your code here
    	return (current!=null || !myStack.isEmpty());
    }
    
    //@return: return next node
    public TreeNode next() {
        // write your code here
    	while(current!=null){
    		myStack.push(current);
    		current = current.left;
    	}   	
    	TreeNode res = (TreeNode) myStack.pop();
    	current = res.right;
       	return res;
    }
}

核心：TreeNode next()

主要功能一直找出最小的那个node：

1. 如果current != null，一直找left.left.left.....就是所求结果；

2. 如果current == null 并且current是 current.parent的left node时，current.parent 就是所求结果，也就是myStack最上面的node；

3. 如果current == null 并且current是 current.parent的right node时， current.parent已经被pop了，current.parent.parent是所求结果，也是myStack最上面的node。

第一个while loop非常巧妙：当current不等于null时，一直查询left node，并把沿途left node存入myStack，一直到最小的node的left node (值为null)，然后再从stack里面读出来上一个left node，虽然可以用while(current.left!=null)来省略两步(current=current.left 和 current=myStack.pop() )，但这样的话current == null的情况要多写一个if语句。while(curerent != null) 包括了 if(current == null) 的情况。之后的从myStack读出来是两个情况都要做的下一步。

current 为结果时，再从以 current.right 为 root 的 tree 中找 next()，所以最后要 current = current.right