A1102. Invert a Binary Tree (25)

题目描述

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

题意解析

题目其实很简单，就是按照样例的输入，构建一颗二叉树，然后依次层次遍历和中序遍历。

解题思路

采用链表存储二叉树，便于层次遍历和中序遍历，关键是根据输入创建一颗二叉树。首先遍历输入，找出根节点（节点编号从0~N，没有出现的就是根节点）。先创建根节点并入队，然后遍历队列，当当前节点有左子树时就创建并入队，有右子数亦然，队列为空时，二叉树就创建好了。最后，按照要求遍历即可。

参考代码

<span style="font-size:14px;">#include <cstdio>#include <queue>#include <malloc.h>using namespace std;const int maxn = 20;int n;bool flag[maxn] = {false};struct N{char l;char r;}temp[maxn];struct node{int data;struct node * lchild;struct node * rchild;};queue<node *> q;node * createTree(int head){node * root = (node *)malloc(sizeof(node));root->data = head;root->lchild = NULL;root->rchild = NULL;q.push(root);while(!q.empty()){node * nroot = q.front();int i = nroot->data;q.pop();if(temp[i].l != '-'){int l = temp[i].l - '0';node * n = (node *)malloc(sizeof(node));n->data = l;n->lchild = NULL;n->rchild = NULL;nroot->lchild = n;q.push(n);}if(temp[i].r != '-'){int r = temp[i].r - '0';node * n = (node *)malloc(sizeof(node));n->data = r;n->lchild = NULL;n->rchild = NULL;nroot->rchild = n;q.push(n);}}return root;}queue<node *> order_q;void level_order(node * root){order_q.push(root);int i = 0;while(!order_q.empty()){node * n = order_q.front();order_q.pop();if(i == 0)printf("%d",n->data);elseprintf(" %d",n->data);i++;if(n->rchild != NULL)order_q.push(n->rchild);if(n->lchild != NULL)order_q.push(n->lchild);}}int x = 0;void in_order(node * root){if(root->rchild != NULL)in_order(root->rchild);if(x == 0){printf("%d",root->data);x++;}else{printf(" %d",root->data);}if(root->lchild != NULL)in_order(root->lchild);}int main(){node * root;int head;scanf("%d",&n);getchar();for(int i=0;i<n;i++){scanf("%c %c",&temp[i].l,&temp[i].r);getchar();}for(int i=0;i<n;i++){if(temp[i].l != '-')flag[temp[i].l-'0'] = true;if(temp[i].r != '-')flag[temp[i].r-'0'] = true;}for(int i=0;i<n;i++){if(flag[i] == false){head = i;break;}}root = createTree(head);level_order(root);printf("\n");in_order(root);return 0;}</span>