題目見這裏

　 和1099略微相似，考察二叉樹和基本的遍歷，算是簡單的啦，下標還充當了數據域，主要是知道要標記訪問到的下標,從而確定root

//1102:Invert a Binary Tree
#include <cstdio>
#include <iostream> 

using namespace std;

const int N = 10;
typedef struct node{
	int lChild,rChild;
}Node;

void Input(int &root, bool *flag, Node *node, int &n){
	int i;
	char left,right;
	scanf("%d",&n);
	getchar();
	for(i=0;i<n;i++){
		scanf("%c %c",&left,&right);
		getchar();
		if(left==‘-‘) node[i].lChild = -1;
		else{
			node[i].lChild = left-‘0‘;
			flag[left-‘0‘] = true;
		}
		if(right==‘-‘) node[i].rChild = -1;
		else{
			node[i].rChild = right-‘0‘;
			flag[right-‘0‘] = true;
		}
	}
	for(i=0;i<n;i++) 
		if(!flag[i]){
			root = i;
			break;
		}  
}

void Invert(int root, Node *node){
	if(root!=-1){
		Invert(node[root].lChild,node);
		Invert(node[root].rChild,node);
		swap(node[root].lChild,node[root].rChild);
	}
}

void LevelOrderTraverse(int root, Node *node, int n){
	int qNode[N],tmpNode;
	int front,rear;
	front = rear = 0;
	if(root==-1) return;
	qNode[rear] = root;
	rear ++;
	while(front<rear){ //無‘=‘ 
		tmpNode = qNode[front];
		front ++;
		printf("%d",tmpNode);
		if(front<n) printf(" ");
		else printf("\n");
		if(node[tmpNode].lChild!=-1){
			qNode[rear]	= node[tmpNode].lChild;
			rear ++;
		}
		if(node[tmpNode].rChild!=-1){
			qNode[rear]	= node[tmpNode].rChild;
			rear ++;
		}
	}
}

void InOrderTraverse(int root, Node *node, int n){
	static int i = 1;
	if(root!=-1){
		InOrderTraverse(node[root].lChild,node,n);
		printf("%d",root);
		if(i==n) printf("\n");
		else printf(" ");
		i ++;
		InOrderTraverse(node[root].rChild,node,n);
	}
}

int main(){
//	freopen("Data.txt","r",stdin);
	Node node[N];
	bool flag[N]={false};
	int root,n;
	Input(root,flag,node,n);
	Invert(root,node);
	LevelOrderTraverse(root,node,n);
	InOrderTraverse(root,node,n);
	return 0;
}

PAT-1102(Invert a Binary Tree)