PAT(A) - 1102. Invert a Binary Tree (25)

1102. Invert a Binary Tree (25)

时间限制

400 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CHEN, Yue

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:

81 -- -0 -2 7- -- -5 -4 6

Sample Output:

3 7 2 6 4 0 5 16 5 7 4 3 2 0 1

思路分析：输入结点建立一棵二叉树，关键先找到根节点，然后进行遍历。输出第一行是层序遍历，第二行是中序遍历。然而。。。输出结果怎么和样例不太一样呢。原来还有一个单词是'Invert"，意思是翻转。不用真正的做翻转，只要遍历的时候调整一下顺序就OK了。

#include <cstdio>#include <queue>#define MAX 10using namespace std;typedef struct {    int index;    int lchild;    int rchild;} Node;Node node[MAX];int count = 0;void dfsInOrder( Node curNode ) {    if( curNode.rchild != -1 ) {        dfsInOrder( node[curNode.rchild] );    }    if( count == 0 ) {       printf( "%d", curNode.index );       count = 1;    }    else {       printf( " %d", curNode.index );    }    if( curNode.lchild != -1 ) {        dfsInOrder( node[curNode.lchild] );    }}int main() {    //freopen( "123.txt", "r", stdin );    int isRoot[MAX] = { 0 };    int n;    scanf( "%d", &n );    char ch = getchar();    for( int i = 0; i < n; i++ ) {        node[i].index = i;        char left, right;        scanf( "%c %c", &left, &right );        char ch = getchar();        if( left == '-' ) node[i].lchild = -1;        else { node[i].lchild = left - '0'; isRoot[left - '0'] = 1; }        if( right == '-' ) node[i].rchild = -1;        else { node[i].rchild = right - '0'; isRoot[right - '0'] = 1; }    }    //for( int i = 0; i < n; i++ ) {    //    printf( "%d %d %d\n", node[i].index, node[i].lchild, node[i].rchild );    //}    int root = 0;    for( int i = 0; i < n; i++ ) {        if( isRoot[i] == 0 ) {           root = i;           break;        }    }    //printf( "root: %d\n", root );    queue<Node> q;    q.push( node[root] );    while( !q.empty() ) {        Node curNode = q.front();        q.pop();        if( count == 0 ) {            printf( "%d", curNode.index );            count = 1;        }        else            printf( " %d", curNode.index );        if( curNode.rchild != -1 ) q.push( node[curNode.rchild] );        if( curNode.lchild != -1 ) q.push( node[curNode.lchild] );    }    printf( "\n" );    count = 0;    dfsInOrder( node[root] );    return 0;}