标签： OJ

The following is from Max Howell @twitter:
Google: 90% of our engineers use the software you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck off.
Now it’s your turn to prove that YOU CAN invert a binary tree!

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=10) which is the total number of nodes in the tree — and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node from 0 to N-1, and gives the indices of the left and right children of the node. If the child does not exist, a “-” will be put at the position. Any pair of children are separated by a space.

Output Specification:

For each test case, print in the first line the level-order, and then in the second line the in-order traversal sequences of the inverted tree. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.

Sample Input:
8
1 –
– –
0 –
2 7
– –
– –
5 –
4 6
Sample Output:
3 7 2 6 4 0 5 1
6 5 7 4 3 2 0 1
题目大意：反转一棵二叉树，给出原二叉树的每个结点的左右孩子，输出它的层序和中序遍历～

分析：1. 反转二叉树就是存储的时候所有左右结点都交换。
2. 二叉树使用{id, l, r, index, level}存储每个结点的id, 左右结点,下标值，和当前层数～
3. 根结点是所有左右结点中没有出现的那个结点～
4. 已知根结点，用递归的方法可以把中序遍历的结果push_back到数组v1里面，直接输出就是中序，排序输出就是层序（排序方式，层数小的排前面，相同层数时，index大的排前面）

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
Both the left and right subtrees must also be binary search trees.
A Complete Binary Tree (CBT) is a tree that is completely filled, with the possible exception of the bottom level, which is filled from left to right.

Now given a sequence of distinct non-negative integer keys, a unique BST can be constructed if it is required that the tree must also be a CBT. You are supposed to output the level order traversal sequence of this BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N distinct non-negative integer keys are given in the next line. All the numbers in a line are separated by a space and are no greater than 2000.

Output Specification:

For each test case, print in one line the level order traversal sequence of the corresponding complete binary search tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input:
10
1 2 3 4 5 6 7 8 9 0
Sample Output:
6 3 8 1 5 7 9 0 2 4
题目大意：给一串构成树的序列，已知该树是完全二叉搜索树，求它的层序遍历的序列
分析：总得概括来说，已知中序，从根节点开始中序遍历，按中序数组给出的顺序依次将值填入level数组对应的下标中，输出level数组可得层序遍历。
1. 因为二叉搜索树的中序满足：是一组序列的从小到大排列，所以只需将所给序列排序即可得到中序数组in
2. 假设把树按从左到右、从上到下的顺序依次编号，根节点为0，则从根结点root = 0开始中序遍历，root结点的左孩子下标是root*2+1，右孩子下标是root*2+2
3. 因为是中序遍历，所以遍历结果与中序数组in中的值从0开始依次递增的结果相同，即in[t++]（t从0开始），将in[t++]赋值给level[root]数组
4. 因为树是按从左到右、从上到下的顺序依次编号的，所以level数组从0到n-1的值即所求的层序遍历的值，输出level数组即可～

Given a tree, you are supposed to tell if it is a complete binary tree.

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=20) which is the total number of nodes in the tree — and hence the nodes are numbered from 0 to N-1. Then N lines follow, each corresponds to a node, and gives the indices of the left and right children of the node. If the child does not exist, a “-” will be put at the position. Any pair of children are separated by a space.

Output Specification:

For each case, print in one line “YES” and the index of the last node if the tree is a complete binary tree, or “NO” and the index of the root if not. There must be exactly one space separating the word and the number.

Sample Input 1:
9
7 8
– –
– –
– –
0 1
2 3
4 5
– –
– –
Sample Output 1:
YES 8
Sample Input 2:
8
– –
4 5
0 6
– –
2 3
– 7
– –
– –
Sample Output 2:
NO 1

题目大意：给出一个n表示有n个结点，这n个结点为0~n-1，给出这n个结点的左右孩子，求问这棵树是不是完全二叉树
分析：递归出最大的下标值，完全二叉树一定把前面的下标充满： 最大的下标值 == 最大的节点数；不完全二叉树前满一定有位置是空，会往后挤： 最大的下标值 > 最大的节点数～

一棵二叉搜索树可被递归地定义为具有下列性质的二叉树：对于任一结点，

其左子树中所有结点的键值小于该结点的键值；
其右子树中所有结点的键值大于等于该结点的键值；
其左右子树都是二叉搜索树。
所谓二叉搜索树的“镜像”，即将所有结点的左右子树对换位置后所得到的树。

给定一个整数键值序列，现请你编写程序，判断这是否是对一棵二叉搜索树或其镜像进行前序遍历的结果。

输入格式：

输入的第一行给出正整数N（<=1000）。随后一行给出N个整数键值，其间以空格分隔。

输出格式：

如果输入序列是对一棵二叉搜索树或其镜像进行前序遍历的结果，则首先在一行中输出“YES”，然后在下一行输出该树后序遍历的结果。数字间有1个空格，一行的首尾不得有多余空格。若答案是否，则输出“NO”。

输入样例1：
7
8 6 5 7 10 8 11
输出样例1：
YES
5 7 6 8 11 10 8
输入样例2：
7
8 10 11 8 6 7 5
输出样例2：
YES
11 8 10 7 5 6 8
输入样例3：
7
8 6 8 5 10 9 11
输出样例3：
NO

分析：假设它是二叉搜索树，一开始isMirror为FALSE，根据二叉搜索树的性质将已知的前序转换为后序，转换过程中，如果发现最后输出的后序数组长度不为n，那就设isMirror为true，然后清空后序数组，重新再转换一次（根据镜面二叉搜索树的性质），如果依旧转换后数组大小不等于n，就输出no否则输出yes

A Binary Search Tree (BST) is recursively defined as a binary tree which has the following properties:

The left subtree of a node contains only nodes with keys less than the node’s key.
The right subtree of a node contains only nodes with keys greater than or equal to the node’s key.
Both the left and right subtrees must also be binary search trees.
If we swap the left and right subtrees of every node, then the resulting tree is called the Mirror Image of a BST.

Now given a sequence of integer keys, you are supposed to tell if it is the preorder traversal sequence of a BST or the mirror image of a BST.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (<=1000). Then N integer keys are given in the next line. All the numbers in a line are separated by a space.

Output Specification:

For each test case, first print in a line “YES” if the sequence is the preorder traversal sequence of a BST or the mirror image of a BST, or “NO” if not. Then if the answer is “YES”, print in the next line the postorder traversal sequence of that tree. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:
7
8 6 5 7 10 8 11
Sample Output 1:
YES
5 7 6 8 11 10 8
Sample Input 2:
7
8 10 11 8 6 7 5
Sample Output 2:
YES
11 8 10 7 5 6 8
Sample Input 3:
7
8 6 8 5 10 9 11
Sample Output 3:
NO
题目大意：给定一个整数键值序列，现请你编写程序，判断这是否是对一棵二叉搜索树或其镜像进行前序遍历的结果。
分析：假设它是二叉搜索树，一开始isMirror为false，根据二叉搜索树的性质将已知的前序转换为后序，转换过程中，如果发现最后输出的后序数组长度不为n，那就设isMirror为true，然后清空后序数组，重新再转换一次（根据镜面二叉搜索树的性质），如果依旧转换后数组大小不等于n，就输出NO否则输出YES