PAT Advanced 1102 Invert a Binary Tree (25) [树的遍历]

题目

The following is from Max Howell @twitter:
Google: 90% of our engineers use the sofware you wrote (Homebrew), but you can’t invert a binary tree on a whiteboard so fuck of. 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 lef 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

题目分析

已知所有节点的左右子节点，求反转二叉树的中序和后序序列

解题思路

思路 01

1. 输入时，将左右子节点对换，即可完成反转
2. bfs广度优先遍历，输出层序序列
3. 递归输出中序序列

思路 02

1. 将节点关系按照输入保存
2. 使用后序遍历递归进行二叉树反转（也可使用前序遍历递归进行二叉树反转）
3. bfs广度优先遍历，输出层序序列
4. 递归输出中序序列

Code

Code 01（最优）

#include <iostream>
#include <vector>
#include <queue>
using namespace std;
const int maxn = 10;
int nds[maxn][2];
int n,cnt;
bool flag[maxn];
void bfs(int root) {
	queue<int> q;
	q.push(root);
	while(!q.empty()) {
		int now = q.front();
		q.pop();
		printf("%d",now);
		if(++cnt<n)printf(" ");
		if(nds[now][0]!=-1)q.push(nds[now][0]);
		if(nds[now][1]!=-1)q.push(nds[now][1]);
	}
}
void inOrder(int nd){
	if(nd==-1){//nds[nd][0]==-1&&nds[nd][1]==-1
		return;
	}
	inOrder(nds[nd][0]);
	printf("%d",nd);
	if(++cnt<n)printf(" ");
	inOrder(nds[nd][1]);
}
int main(int argc,char * argv[]) {
	char f,r;
	scanf("%d",&n);
	for(int i=0; i<n; i++) {
		scanf("%*c%c %c",&r,&f);
		if(f=='-')nds[i][0]=-1;
		else {
			nds[i][0]=f-'0';
			flag[nds[i][0]]=true;
		}
		if(r=='-')nds[i][1]=-1;
		else {
			nds[i][1]=r-'0';
			flag[nds[i][1]]=true;
		}

	}
	//find root
	int k=0;
	while(k<n&&flag[k])k++;
	bfs(k);
	printf("\n");
	cnt=0;
	inOrder(k);
	return 0;
}

Code 02

#include <cstdio>
#include <queue>
#include <algorithm>
using namespace std;
const int maxn = 10;
struct node {    // 二叉树的静态写法
    int lchild, rchild;
} Node[maxn];
bool notRoot[maxn] = {false};    // 记录是否不是根结点，初始均是根结点
int n, num = 0;    // n为结点个数，num为当前已经输出的结点个数
// print函数输出结点id的编号
void print(int id) {
    printf("%d", id);    // 输出id
    num++;    // 已经输出的结点个数加1
    if(num < n) printf(" ");    // 最后一个结点不输出空格
    else printf("\n");
}
// 中序遍历
void inOrder(int root) {
    if(root == -1) {
        return;
    }
    inOrder(Node[root].lchild);
    print(root);
    inOrder(Node[root].rchild);
}
// 层序遍历
void BFS(int root) {
    queue<int> q;  //注意队列里是存地址
    q.push(root);  //将根结点地址入队
    while(!q.empty()) {
        int now = q.front();  //取出队首元素
        q.pop();
        print(now);
        if(Node[now].lchild != -1) q.push(Node[now].lchild);  //左子树非空
        if(Node[now].rchild != -1) q.push(Node[now].rchild);  //右子树非空
    }
}
// 后序遍历，用以反转二叉树
//void postOrder(int root) {
//    if(root == -1) {
//        return;
//    }
//    postOrder(Node[root].lchild);
//    postOrder(Node[root].rchild);
//    swap(Node[root].lchild, Node[root].rchild);    // 交换左右孩子
//}
// 前序遍历，用以反转二叉树
void preOrder(int root) {
    if(root == -1) {
        return;
    }
    swap(Node[root].lchild, Node[root].rchild);    // 交换左右孩子
    preOrder(Node[root].lchild);
    preOrder(Node[root].rchild);
}
// 将输入的字符转换为-1或者结点编号
int strToNum(char c) {
    if(c == '-') return -1;    // “-”表示没有孩子结点，记为-1
    else {
        notRoot[c - '0'] = true;    // 标记c不是根结点
        return c - '0';    // 返回结点编号
    }
}
// 寻找根结点编号
int findRoot() {
    for(int i = 0; i < n; i++) {
        if(notRoot[i] == false) {
            return i;    // 是根结点，返回i
        }
    }
}
int main() {
    char lchild, rchild;
    scanf("%d", &n);    // 结点个数
    for(int i = 0; i < n; i++) {
        scanf("%*c%c %c", &lchild, &rchild);    // 左右孩子
        Node[i].lchild = strToNum(lchild);
        Node[i].rchild = strToNum(rchild);
    }
    int root = findRoot();    // 获得根结点编号
//    postOrder(root);    // 后序遍历，反转二叉树
    preOrder(root);    // 前序遍历，反转二叉树
    BFS(root);    // 输出层序遍历序列
    num = 0;    // 已输出的结点个数置0
    inOrder(root);    // 输出中序遍历序列
    return 0;
}