Home IT Linux Windows Database Network Programming Server Mobile  
  Home \ Programming \ List Leaves     - SSH configuration under Linux (Linux)

- IOS distributed management using GitHub project development (Linux)

- Nagios plugin installation tutorial of Nrpe (Linux)

- ActiveMQ configuration Getting Started Tutorial (Server)

- Daemon under Linux (Linux)

- C # mobile side and PC-side data exchange (Database)

- Compile and install Ubuntu Linux 4.0.5 kernel, network and fix vmware kernel module compilation error (Linux)

- Migu online music player for Linux (Linux)

- CentOS 6.4 installation environment to build Scrapy 0.22 (Linux)

- Elasticsearch Kibana installation notes (Linux)

- Win7 + Ubuntu Kylin + CentOS 6.5 installed three systems (Linux)

- Windows 8.1 and Ubuntu 14.04 dual system uninstall Ubuntu Tutorial (Linux)

- C # using the HttpClient Note: Preheat the long connection (Programming)

- How to set cache valid time in Apache (Server)

- MongoDB 3.2 Cluster Setup (Database)

- How to configure chroot environment in Ubuntu 14.04 (Linux)

- Simple Linux file system - df, du, ln (Linux)

- Java memory-mapped file MappedByteBuffer (Programming)

- How to use the process on the desktop xkill end Linux (Linux)

- OpenGL Superb Learning Notes - Vertex Shader example (Programming)

  List Leaves
  Add Date : 2017-04-13      
  Given a tree, you are supposed to list all the leaves in the order of top down, and left to right.

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, 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 one line all the leaves' indices in the order of top down, and left to right. There must be exactly one space between any adjacent numbers, and no extra space at the end of the line.

Sample Input:

1 -
- -
0 -
- -
- -
5 -
Sample Output:


Title to the effect: By entering the number of nodes and the left son and right son of each node, leaf nodes are printed upside down.
Key topics: To understand the first few lines of the input node is to represent the value of a few. Such as sample input in the 0th row 1 - Represents the value of left child node 0 is 1, which points to the first line, the right child is empty (-1)

Code is as follows:

#define N 10

typedef struct Node
    int data, left, right;
} TreeNode;
TreeNode node [N];
TreeNode Queue [N]; // array queue

int first = -1, last = -1;

void Push (TreeNode tn);
TreeNode Pop ();
void printLeaves (int root, int n);

int charToInt (char ch);

int main ()
    int n;
    bool isRoot [N];
    int root;

    scanf ( "% d \ n", & n);
    for (int i = 0; i         isRoot [i] = 1;
    for (int i = 0; i     {
        char cLeft, cRight;
        scanf ( "% c% c", & cLeft, & cRight);
        getchar (); // read carriage buffer zone
        node [i] .left = charToInt (cLeft);
        node [i] .right = charToInt (cRight);
        node [i] .data = i;
        // Node left child and a right child is certainly not the root
        if (node [i] .left! = -1)
            isRoot [node [i] .left] = 0;
        if (node [i] .right! = -1)
            isRoot [node [i] .right] = 0;
    // Find the root
    for (int i = 0; i     {
        if (isRoot [i])
            root = i;
    printLeaves (root, n);

    return 0;

void Push (TreeNode treeNode)
    Queue [++ last] = treeNode;

TreeNode Pop ()
    return Queue [++ first];

// Node tree traversal sequence and print out a leaf node: queue implementation
void printLeaves (int root, int n)
    int leaves [N];
    int k = 0;
    Push (node [root]);
    for (int i = 0; i     {
        TreeNode tn = Pop ();
        // The left child and right child does not exist, it will save the value of the leaf node to the array, formatted for easy printing
        if (tn.left == -1 && tn.right == -1)
            leaves [k ++] = tn.data;
        if (tn.left! = -1)
            Push (node [tn.left]);
        if (tn.right! = -1)
            Push (node [tn.right]);
    for (int i = 0; i         printf ( "% d", leaves [i]);
    printf ( "% d \ n", leaves [k-1]);

int charToInt (char ch)
    if (isdigit (ch))
        return ch - '0';
        return -1;
- PostgreSQL 9.3.2 Json type of use (Database)
- Ubuntu users to install voice switch instructs the applet (Linux)
- Java implementation of stacks and queues (Programming)
- Ubuntu Linux use ufw or iptables firewall configuration (Linux)
- Git uses a small mind (Linux)
- Install and configure GO 1.2.1 under CentOS 6.5 (Linux)
- ctop: monitor container performance Linux command line artifact (Linux)
- Proper use Core Data multithreaded 3 ways (Programming)
- Setting grep Highlight Matches (Linux)
- Grading defense against Linux server attacks (Linux)
- UNIX how to restrict users by IP Telnet (Linux)
- Ease of use "Explain Shell" script to understand Shell command (Linux)
- Linux supports serial output method (Linux)
- Nginx Module Development - get user ip (Server)
- Linux command line to put on your coat GUI (Linux)
- Configuration based on open source Lucene Java development environment (Server)
- Automated Password Generator: Linux under a special password generator (Linux)
- Oracle 11g on Linux system boot from the startup settings (Database)
- Swift 2.0 brief (Linux)
- Python script running in the background (Programming)
  CopyRight 2002-2016 newfreesoft.com, All Rights Reserved.