【100题】二叉树中找出和为某一值的所有路径

//在二叉树中找出和为某一值的所有路径
#include <iostream>
#include <vector>
using namespace std;

//树结构
struct BTreeNode
{
	int m_nValue;
	BTreeNode *m_pLeft;
	BTreeNode *m_pRight;
}*BiTree;

//按照先序创建二叉树,0表示空
BTreeNode *Create()
{
    int ch;
    cin>> ch;
    if(ch == 0)
    {
         return NULL;
    }
    else
    {
         BTreeNode *root = new BTreeNode();
		 root->m_nValue = ch;
		 root->m_pLeft = Create();
		 root->m_pRight = Create();
         return root;
    }
}

//按照先序 打印 树
void printBTree(BTreeNode *root)
{
	if(!root)
	{
		return ;
	}
	cout << root->m_nValue <<" ";
	printBTree(root->m_pLeft);
	printBTree(root->m_pRight);
}

//查找路径
void FindPath(BTreeNode *pTreeNode,int expectedSum,vector<int> &path,int ¤tSum)
{
	if(!pTreeNode)
	{
		return ;
	}

	//累加到和里
	currentSum += pTreeNode->m_nValue;
	//将该节点的数据域入栈
	path.push_back(pTreeNode->m_nValue);

	bool isLeaf = (!pTreeNode->m_pLeft && !pTreeNode->m_pRight);
	//若当前为叶子节点，并且和恰好达到，则输出
	if(currentSum == expectedSum && isLeaf)
	{
		for(vector<int>::iterator iter = path.begin(); iter != path.end(); ++iter)
		{
			cout << *iter << '\t';
		}
		cout << endl;
	}
	//若不是叶子节点，则或者有左孩子
	if(pTreeNode->m_pLeft)
	{
		FindPath(pTreeNode->m_pLeft,expectedSum,path,currentSum);
	}
	//或者有右孩子
	if(pTreeNode->m_pRight)
	{
		FindPath(pTreeNode->m_pRight,expectedSum,path,currentSum);
	}
	//
	currentSum -= pTreeNode->m_nValue;
	path.pop_back();
	return ;
}

void main()
{
	BTreeNode *root=NULL;
	cout << "创建树："<<endl;
	root = Create();
	cout << "打印树："<<endl;
	printBTree(root);
	vector<int> temp;
	int b = 0;
	int expectedSum;
	int &a = b;
	cout <<endl<<"请输入期望的和：";
	cin >> expectedSum;
	cout << "满足条件的路径有："<<endl;
	FindPath(root,expectedSum,temp,a);
}

运行结果：

 

创建树：
1 3 4 0 0 5 0 0 2 6 0 0 7 0 0
打印树：
1 3 4 5 2 6 7
请输入期望的和：9
满足条件的路径有：
1       3       5
1       2       6
请按任意键继续. . .