【PAT】1102. Invert a Binary Tree (25)

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 1
6 5 7 4 3 2 0 1
题意：将二叉树反转（即左右节点交换），然后给出层序遍历和中序遍历的结果。
分析：层序遍历用队列即可，中序遍历用栈来实现。不难，但是要仔细。
#include <iostream>#include <vector>#include <queue>#include <stack>using namespace std;struct node{node(){left = -1;right = -1;father = -1;level = 0;}int index;int left;int right;int father;int level;};int main(int argc, char** argv) {int n, i;scanf("%d",&n);getchar();vector<node> tree(n);char L, R;int left , right;for(i=0; i<n; i++){tree[i].index = i;scanf("%c %c",&L, &R);getchar();if(isdigit(L)){left = L - '0';tree[i].right = left ;tree[left].father = i;}if(isdigit(R)){right = R - '0';tree[i].left = right;tree[right].father = i;}}int root;for(i=0; i<n; i++){if(tree[i].father == -1){ root = i;break;}}queue<int> que;int level, index;que.push(root);printf("%d",root);bool first = false;node tmpNode;while(!que.empty()){tmpNode = tree[que.front()];if(!first){first = true;}else{printf(" %d",que.front());}level = tmpNode.level;left = tmpNode.left;right = tmpNode.right;if(left != -1){que.push(left);}if(right != -1){que.push(right);}que.pop();}printf("\n");stack<int> sta; sta.push(root);  int cnt = 0; while(!sta.empty()){ index = sta.top();tmpNode = tree[index];left = tmpNode.left;while(left != -1){tree[index].left = -1;sta.push(left);tmpNode = tree[left];index = tmpNode.index;left = tmpNode.left; }  cnt++; if(cnt==n){ printf("%d\n",sta.top());}else{printf("%d ",sta.top());}sta.pop();    if(tmpNode.right != -1){    sta.push(tmpNode.right);} }return 0;}

另一种写法：
#include <iostream>#include <vector>#include <queue>using namespace std;struct node{node(){fa = -1;left = -1;right = -1;}int id, fa, left, right;};vector<node> tree;void inOrder(int r, vector<int> &vec){if(tree[r].right != -1){inOrder(tree[r].right,vec);}vec.push_back(r);if(tree[r].left!=-1){inOrder(tree[r].left,vec);}}void levelOrder(int r, vector<int> &vec){queue<node> que;que.push(tree[r]);while(!que.empty()){node n = que.front();if(n.right!= -1){que.push(tree[n.right]);}if(n.left!=-1){que.push(tree[n.left]);}vec.push_back(n.id);que.pop();}}void show(vector<int> &vec){int i;for(i=0; i<vec.size(); i++){if(i==0){printf("%d",vec[i]);}else{printf(" %d",vec[i]);}}printf("\n");}int main(int argc, char** argv) {int N;scanf("%d",&N);getchar();int i;tree.resize(N);char a, b;for(i=0; i<N; i++){tree[i].id = i;scanf("%c %c",&a,&b);getchar();if(isdigit(a)){tree[a-'0'].fa = i;tree[i].left = a-'0';}if(isdigit(b)){tree[b -'0'].fa = i;tree[i].right = b-'0';}}int root;for(i=0; i<N; i++){if(tree[i].fa==-1){root = i;break;}}vector<int> level;levelOrder(root, level);show(level);vector<int> vec;inOrder(root,vec);show(vec);return 0;}