PAT甲级Invert a Binary Tree 柳神层序遍历的思路值得借鉴

1102 Invert a Binary Tree (25分)

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

思路：就注意是翻转的二叉树，反转就是输入的左孩子，保存到右孩子的位置，就行了，然后层序中序遍历就好啦。

#include <iostream>
#include <string>
using namespace std;
#include<vector>
#include<queue>struct node
{int num;node *rchild,*lchild;node():rchild(NULL),lchild(NULL) {}//最初时候左右孩子节点指针都是NULL
};vector<int> out2;
vector<int> out1;void inOrder(node *root)//中序遍历
{if(root==NULL)return;inOrder(root->lchild);//cout<<root->num;out2.push_back(root->num);inOrder(root->rchild);
}int main()
{int N;cin>>N;int info[N][2];for(int i=0; i<N; i++)//记录保存结点左右孩子的信息{char a,b;cin>>a>>b;if(a!='-'){info[i][0]=a-'0';}elseinfo[i][0]=-1;if(b!='-'){info[i][1]=b-'0';}elseinfo[i][1]=-1;}int sign[N]= {0}; //这样才能全部赋值为0  ,里面填1 就是sign[0]是1 其他为0，求那个事根节点node sn[N];//从新开辟了空间g++ 编译器 可支持这种写法，要不然就直接生成999个结点的数组也行for(int i=0; i<N; i++)//赋值每个孩子结点的左右指针，并顺便判断谁是根节点，没有父节点的就是了{sn[i].num=i;if(info[i][1]!=-1)//1存储的右孩子信息，但是使用lchild 左孩子指针指向，这就是翻转了{sn[i].lchild=&sn[info[i][1]];sign[info[i][1]]=-1;}if(info[i][0]!=-1){sn[i].rchild=&sn[info[i][0]];sign[info[i][0]]=-1;}}int root;for(int i=0; i<N; i++)if(sign[i]==0)root=i;inOrder(&sn[root]);//下面是实现层序遍历vector<int> out1;queue<node*> que;que.push(&sn[root]);while(que.size()!=0){queue<node*> get;while(!que.empty()){out1.push_back(que.front()->num);if(que.front()->lchild!=NULL)get.push(que.front()->lchild);if(que.front()->rchild!=NULL)get.push(que.front()->rchild);que.pop();}que=get;}//层序输出for(int i=0; i<out1.size(); i++){if(i==0)cout<<out1[i];elsecout<<' '<<out1[i];}//中序输出cout<<endl;for(int i=0; i<out2.size(); i++){if(i==0)cout<<out2[i];elsecout<<' '<<out2[i];}return 0;
}

我也看了看柳神的代码，感觉还是差距啊，50行搞定，层序遍历实现也很巧妙

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PAT甲级Invert a Binary Tree 柳神层序遍历的思路值得借鉴

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