剑指offer面试题25-二叉树中和韦某一值的路径

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question：给定一颗二叉树，一个值。我们希望输出和为该值的所有路径。

假设树的结构如下

思路：使用一个基于先序遍历的递归过程来遍历树中每个点一次，每遍历一个点就要获得从根节点到被访问的点的累加和，并保存从根节点到当前点的路径。然后进行判断，

case1如果当前点的左右孩子都为空（叶子节点）并且累加和等于我们期望的和，那么就打印出路径；

case2如果左孩子为空，递归到最孩子；

case3如果右孩子为空，递归到右孩子；

case4如果是叶子并且当前和不等于期望和，网上回退

试下代码：

参数为值传递的代码

#include<vector>#include<iostream>using namespace std;typedef struct BinaryTreeNode{int data;struct BinaryTreeNode *lchild;struct BinaryTreeNode *rchild;}*BinTreeRoot;void createTree1(BinaryTreeNode *&root){root = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));root->data = 8;root->lchild = NULL;root->rchild = NULL;BinaryTreeNode *p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 8;p->lchild = NULL;p->rchild = NULL;root->lchild = p;p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 7;p->lchild = NULL;p->rchild = NULL;root->rchild = p;p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 9;p->lchild = NULL;p->rchild = NULL;root->lchild->lchild = p;p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 2;p->lchild = NULL;p->rchild = NULL;root->lchild->rchild = p;p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 4;p->lchild = NULL;p->rchild = NULL;root->lchild->rchild->lchild = p;p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 7;p->lchild = NULL;p->rchild = NULL;root->lchild->rchild->rchild = p;}void findPath(BinaryTreeNode *root, int expectedSum, int curSum, vector<int>path){if (root == NULL)return;curSum += root->data;path.push_back(root->data);if (root->lchild == NULL && root->rchild == NULL && curSum == expectedSum){for (auto iter = path.begin(); iter != path.end();  ++iter){if (iter + 1 == path.end()){cout << *iter << endl;}else{cout << *iter << "->";}}}if (root->lchild != NULL){findPath(root->lchild, expectedSum, curSum, path);}if (root->rchild != NULL){findPath(root->rchild, expectedSum, curSum, path);}//curSum -= root->data;//path.pop_back();}int main(){BinaryTreeNode *root = NULL;createTree1(root);if (root == NULL){cout << "the tree is empty!" << endl;return 0;}int curSum = 0;//存放从根节点到当前节点的路径和int expectedSum;//我们期待的值cin >> expectedSum;vector<int>path;//存放路径findPath(root, expectedSum, curSum, path);return 0;}

注意：在这个代码中存放累加和curSum、路径path的变量都是使用的值传递，而不是引用传递。因为该方法是一个递归的过程，在每一个递归栈中都有一个临时变量curSum和path。如果递归返回到上一层时，本次计算的没有影响。相反，如果使用了引用传递的，需要手动修改数据的值：

参数为引用传递的代码：

#include<vector>#include<iostream>using namespace std;typedef struct BinaryTreeNode{int data;struct BinaryTreeNode *lchild;struct BinaryTreeNode *rchild;}*BinTreeRoot;void createTree1(BinaryTreeNode *&root){root = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));root->data = 8;root->lchild = NULL;root->rchild = NULL;BinaryTreeNode *p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 8;p->lchild = NULL;p->rchild = NULL;root->lchild = p;p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 7;p->lchild = NULL;p->rchild = NULL;root->rchild = p;p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 9;p->lchild = NULL;p->rchild = NULL;root->lchild->lchild = p;p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 2;p->lchild = NULL;p->rchild = NULL;root->lchild->rchild = p;p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 4;p->lchild = NULL;p->rchild = NULL;root->lchild->rchild->lchild = p;p = (BinaryTreeNode *)malloc(sizeof(BinaryTreeNode));p->data = 7;p->lchild = NULL;p->rchild = NULL;root->lchild->rchild->rchild = p;}void findPath(BinaryTreeNode *root, int expectedSum, int &curSum, vector<int>&path)//引用传递{if (root == NULL)return;curSum += root->data;path.push_back(root->data);if (root->lchild == NULL && root->rchild == NULL && curSum == expectedSum){for (auto iter = path.begin(); iter != path.end();  ++iter){if (iter + 1 == path.end()){cout << *iter << endl;}else{cout << *iter << "->";}}}if (root->lchild != NULL){findPath(root->lchild, expectedSum, curSum, path);}if (root->rchild != NULL){findPath(root->rchild, expectedSum, curSum, path);}curSum -= root->data;//curSum和path是引用传递，所以如果递归返回到上一层的时候，需要减去本次的累加path.pop_back();}int main(){BinaryTreeNode *root = NULL;createTree1(root);if (root == NULL){cout << "the tree is empty!" << endl;return 0;}int curSum = 0;//存放从根节点到当前节点的路径和int expectedSum;//我们期待的值cin >> expectedSum;vector<int>path;//存放路径findPath(root, expectedSum, curSum, path);return 0;}

过程：假设我们希望输出和为25的所有路径；

设定以下变量：int curSum = 0;//存放当前累加和

int expectedSum = 25;//期待和

vector<int>path;//用于存放路径

BinaryTreeNode *root;//指向树中根节点的指针

（1）首先从根节点出发，curSum = 8,并且不是叶子节点，并且值不为35。