归档 2015年1月

给定一棵二叉树，不使用递归，迭代地后序遍历并输出树中的元素

 二叉树的后序遍历很容易采用递归方式实现：

void postOrderTraversal(BinaryTree *p) {
  if (!p) return;
  postOrderTraversal(p->left);
  postOrderTraversal(p->right);
  cout << p->data;
}

后序遍历是二叉树三种遍历的非递归算法中最难实现的一种，在做这道题目之前可以首先尝试一下这一题，因为它相对简单一些：Binary Search Tree In-Order Traversal Iterative Solution。

三种遍历的非递归实现中最容易的是先序遍历。

后序遍历是一种非常有用的树操作，例如它可以被用于下面的场景：

树的删除。为了释放树结构的内存，某节点在被释放以前，其左右子树的节点首先应当被释放掉。
后缀表示法（逆波兰表示法）

如果在遍历树时维护一个visited标记，问题可以比较直观地解决。在此不详细讨论该方法 ...

题目描述：

Given a binary tree, return the inorder traversal of its nodes' values.

For example:
Given binary tree {1,#,2,3},
   1
    \
     2
    /
   3
return [1,3,2].

Note: Recursive solution is trivial, could you do it iteratively?

题目大意：

非递归实现二叉树的中序遍历 ...

题目描述：

John is standing at the origin of an infinite two-dimensional grid. He is going to move along this grid. During each second he can either stay where he is or he can move by one unit in one of ...

题目描述：

Write a SQL query to find all duplicate emails in a table named Person.

+----+---------+
| Id | Email   |
+----+---------+
| 1  | a@b.com |
| 2  | c@d.com |
| 3  | a@b.com |
+----+---------+
For example, your query should return the following for the above ...

题目描述：

The Employee table holds all employees including their managers. Every employee has an Id, and there is also a column for the manager Id.

+----+-------+--------+-----------+
| Id | Name  | Salary | ManagerId |
+----+-------+--------+-----------+
| 1  | Joe   | 70000  | 3         |
| 2  | Henry | 80000  | 4         |
| 3  | Sam   | 60000 ...