PAT甲级——1102 Invert a Binary Tree (层序遍历+中序遍历)

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

题目大意：将二叉树每个节点的左右孩子交换位置，然后分别层序和中序输出。第 i 行代表了节点 i 的左右孩子的信息，先左后右，'-' 表示没有该方向的子节点。

思路：用数组tree存储树。

读取字符串然后将其转成int型变量，空节点'-' 用-1代替。

只要在读取数据的时候交换左右孩子的位置就行。

读取数据的时候用bool数组R对每个孩子节点进行标记，然后遍历数组寻找根节点（即没有被标记过的节点）

#include <iostream> #include <vector> #include <string> #include <queue> using namespace std; struct node { int left, right; }; vector <node> tree; bool flag = false;//用于中序遍历标记第一个输出的节点 int getNum(string &s); void levelOrder(int t); void inOrder(int t); int main() { int N, root; scanf("%d", &N); tree.resize(N); vector <bool> R(N, true); for (int i = 0; i < N; i++) { string left, right; cin >> left >> right; tree[i].left = getNum(right); tree[i].right = getNum(left); if (tree[i].left != -1) R[tree[i].left] = false; if (tree[i].right != -1) R[tree[i].right] = false; } for (int i = 0; i < N; i++) if (R[i]) { root = i; break; } levelOrder(root); printf("\n"); inOrder(root); printf("\n"); return 0; } void inOrder(int t) { if (t != -1) { inOrder(tree[t].left); if (flag) printf(" "); if (!flag) flag = true; printf("%d", t); inOrder(tree[t].right); } } void levelOrder(int t) { queue <int> Q; Q.push(t); while (!Q.empty()) { t = Q.front(); printf("%d", t); Q.pop(); if (tree[t].left != -1) { Q.push(tree[t].left); } if (tree[t].right != -1) { Q.push(tree[t].right); } if (!Q.empty()) printf(" "); } } int getNum(string& s) { if (s[0] == '-') return -1; int n = 0; for (int i = 0; i < s.length(); i++) n = n * 10 + s[i] - '0'; return n; }