标签：tmp Binary right int Invert start 1102 que root

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('\n');
    inorder(root_index);
    return 0;
}

标签：tmp,Binary,right,int,Invert,start,1102,que,root
来源： https://www.cnblogs.com/littlepage/p/12900539.html

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