1102 Invert a Binary Tree
题目:
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
题目大意:根据输入构建二叉树,求出一个倒转二叉树(左右子树交换)的层序和中序遍历结果。
注意: inverted binary tree: 倒转二叉树(左右子树交换)
代码:(中序遍历 + 层序遍历)
#include<stdio.h> #include<iostream> #include<queue> using namespace std; int n, root; bool isChild[12], f = 0; struct Node{ int lchild, rchild; }node[12]; void levelorder(int x){ bool flag = false; queue<int> q; q.push(x); while(!q.empty()){ int tmp = q.front(); q.pop(); if(!flag){ flag = true; }else{ cout<<" "; } cout<<tmp; if(node[tmp].rchild != -1){ q.push(node[tmp].rchild); } if(node[tmp].lchild != -1){ q.push(node[tmp].lchild); } } } void inorder(int x){ if(node[x].rchild != -1){ inorder(node[x].rchild); } if(!f){ f = 1; }else{ cout<<" "; } cout<<x; if(node[x].lchild != -1){ inorder(node[x].lchild); } } int main(){ scanf("%d", &n); for(int i = 0; i < n; i++){ char l, r; cin>>l>>r; // scanf("%c%c", &l, &r); if(l != '-'){ node[i].lchild = l - '0'; isChild[l - '0'] = 1; }else{ node[i].lchild = -1; } if(r != '-'){ node[i].rchild = r - '0'; isChild[r - '0'] = 1; }else{ node[i].rchild = -1; } } for(int i = 0 ; i < n; i++){ if(isChild[i]==0){ root = i; break; } } levelorder(root); cout<<endl; inorder(root); return 0; }
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