1004. Counting Leaves (30)

pat。

这个题目主要就是存储树以及遍历树。

解题心得：

1、可能对c++不是很熟，所以一开始我就循序渐进而用了教材上所给的结构体+指针来写。不过，确实有点晕。

2、我的树存储方法是：孩子存储法，也就是把每个节点都存在一个数组，然而后面链接着他的所有儿子。

3、BFS的原理就是树的层次遍历，所以我还是用了层次遍历，中间再加了一些题目所需要的语句。

4、需要注意的地方：

①：结构体指针用之前需要为它申请内存空间。。。。

②：需要手动把链表最后一个节点置空，否则虽然vs显示无法计算内存地址，但它还是会走到哪儿，然后终止程序。

③：由于我的存储方法是孩子存储法，所以后面循环的不是tree[myParent]，而是他的第一个儿子。

		struct node *firstChild;
		firstChild = (struct node*)malloc(sizeof(struct node));
		firstChild = tree[myParent].firstChild;
		while(NULL != firstChild){
			firstChild->level = level+1;//deeper and increment
			myQueue.push(firstChild);
			firstChild = firstChild->nextChild;
		}

代码如下：

#include"stdio.h"
#include"string.h"
#include"math.h"
#include<iostream>
#include<vector>
#include<queue>
using namespace std;
#define INF 0x7FFFFFFF
//the storage structure of tree,you can choose other ways to store,also.
typedef struct node{
	int value;
	int level;
	struct node *nextChild;
}*Node;
struct PTree{
	int parent;
	struct node *firstChild;
};
vector<struct PTree> tree;
vector<int> nonLeafs;
queue<Node> myQueue;
//level traversal,that is BFS
void levelOrder(){
	int leaf = 0;
	struct node *parent;
	//get the first node and push it into the queue.
	parent = (struct node*)malloc(sizeof(struct node));
	int currentStep = 0;
	parent->value = tree[1].parent;
	parent->nextChild = tree[1].firstChild;
	parent->level = 0;
	myQueue.push(parent);
	while(!myQueue.empty()){   //loop if not empty
		struct node *tempNode;
		tempNode = (struct node*)malloc(sizeof(struct node));
		tempNode = myQueue.front();
		myQueue.pop();
		//get the first node
		int myParent = tempNode->value;
		int level = tempNode->level;
		//modify the level and leaf number
		if(level > currentStep){
			currentStep = level;
			nonLeafs.push_back(leaf);
			leaf = 0;
		}
		//modify key of leaf if it has no child.
		if(tree[myParent].firstChild == NULL){
			leaf++;
		}
		//get the second node and push it into the queue,not the first!
		struct node *firstChild;
		firstChild = (struct node*)malloc(sizeof(struct node));
		firstChild = tree[myParent].firstChild;
		while(NULL != firstChild){
			firstChild->level = level+1;//deeper and increment
			myQueue.push(firstChild);
			firstChild = firstChild->nextChild;
		}
	}
	//add the last node's leaf number.
	nonLeafs.push_back(leaf);
}

int main(){
	int nodes,nonLeafNodes;
	while(scanf("%d%d",&nodes,&nonLeafNodes) != EOF){
		tree.resize(nodes+1);     //size from 1~nodes
		for(int i = 1;i < nonLeafNodes+1;i++){
			int parentNode,nodesSize;
			scanf("%d%d",&parentNode,&nodesSize);
			tree[parentNode].parent = parentNode;
			struct node *tempNode;
			tempNode = (struct node*)malloc(sizeof(struct node));
			tree[parentNode].firstChild = NULL;
			
			for(int j = 0;j < nodesSize;j++){
				// concat the tree.
				int tempChild;
				scanf("%d",&tempChild);
				if(j==0){
					tempNode->value = tempChild;
					tree[parentNode].firstChild = tempNode;
					if(j == nodesSize -1){
						tempNode->nextChild = NULL;
					}
					continue;
				}
				struct node *temp2Node;
				temp2Node = (struct node*)malloc(sizeof(struct node));
				temp2Node->value = tempChild;
				tempNode->nextChild = temp2Node;
				tempNode = temp2Node;
				if(j == nodesSize -1){
					tempNode->nextChild = NULL;
				}
			}
		}
		levelOrder();   //traversal
		//print out
		for(int i = 0;i < nonLeafs.size();i++){
			if(i == nonLeafs.size()-1){
				printf("%d",nonLeafs[i]);
			}else{
				printf("%d ",nonLeafs[i]);
			}

		}
	}
	return 0;
}

当然我觉得作为acm，其实可以不用用指针，网上看到一位大牛，然后他的方法并没有用链表，而是通过c++的stl库中工具完成，我觉得比我的好。

http://download.youkuaiyun.com/detail/anla_/9728778