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 (≤) 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
反转二叉树,这题我的解法是:直接左子树插入右子树,右子树插入左子树,然后就ok啦,之后我们进行层序和中序即可。
#include "iostream" #include "vector" #include "string" #include "queue" using namespace std; int N; string tmp_l, tmp_r; struct node { int data, left = -1, right = -1, level; }; vector<node> v; void levelorder(int root) { bool start = true; queue<node> que; que.push(v[root]); while(!que.empty()) { node n = que.front(); if(start) { printf("%d", n.data); start = false; } else printf(" %d", n.data); que.pop(); if(n.left != -1) que.push(v[n.left]); if(n.right != -1) que.push(v[n.right]); } } bool start = true; void inorder(int root) { if(v[root].left != -1) inorder(v[root].left); if(start) { start = !start; printf("%d", v[root].data); } else printf(" %d", v[root].data); if(v[root].right != -1) inorder(v[root].right); } int main() { scanf("%d", &N); v.resize(N); vector<bool> judge(N, false); for(int i = 0;i < N; i++) { cin >> tmp_l >> tmp_r; v[i].data = i; if(tmp_l != "-") { v[i].right = stoi(tmp_l); judge[stoi(tmp_l)] = true; } if(tmp_r != "-") { v[i].left = stoi(tmp_r); judge[stoi(tmp_r)] = true; } } int root_index; for(int i = 0; i < N; i++) if(judge[i] == false) root_index = i; levelorder(root_index); putchar(' '); inorder(root_index); return 0; }