Cracking The Coding Interview4.8

//You are given a binary tree in which each node contains a value. Design an algorithm to print all paths which sum up to that value. Note that it can be any path in the tree - it does not have to start at the root.//// 译文：//// 给定一棵二叉树，每个结点包含一个值。打印出所有满足以下条件的路径： 路径上结点的值加起来等于给定的一个值。注意：这些路径不必从根结点开始。#include <iostream>#include <string>#include <vector>#include <stack>#include <stdlib.h>using namespace std;class tree{private:struct treenode{char data;treenode * left;treenode * right;treenode * parent;};treenode *root;int index;void create(treenode **p, char *str, int i, int size, treenode *fa){if (i>size-1 || str[i] == '\0'){*p = NULL;return;}if (str[i] == '#'){*p=NULL;}else{*p = new treenode;(*p)->data = str[i];(*p)->parent = fa;create(&((*p)->left),str,2*i+1,size, *p);create(&((*p)->right),str,2*i+2,size, *p);}}void pOrder(treenode *p){if (p==NULL){return;}//cout<<p->data<<"  "<<endl;pOrder(p->left);pOrder(p->right);}void zOrder(treenode *p){if (p==NULL){return;}zOrder(p->left);cout<<p->data<<"  "<<endl;zOrder(p->right);}void hOrder(treenode *p){if (p==NULL){return;}hOrder(p->left);cout<<p->data<<"  "<<endl;hOrder(p->right);}/***遍历每一个节点，朝上找父节点，如果能得到相等的sum ,则将该节点传递给print()***/void findpath(treenode *p, int sum){//cout<<"findpath"<<endl;if (p == NULL){return;}int i = 0;int temp = 0;treenode *pi = p;while(pi != NULL){temp +=pi->data - '0';if (temp == sum){print(p, i);}pi = pi->parent;i++;}findpath(p->left, sum);findpath(p->right, sum);}/***从节点朝父节点找，应反向打印***/void print(treenode *h, int level){if (h == NULL){return ;}//cout<<"print------"<<level<<endl;stack<int>s;int i=0;while(i <= level && h != NULL){s.push(h->data - '0');h = h->parent;i++;}while(!s.empty()){cout<<s.top()<<"  ";s.pop();}cout<<endl;}/***与上一方法类似，只是这里通过vector来记录每个节点上一层经过的路径***/void findpath(treenode *p, int sum, vector<int>&v, int level){if (p == NULL){return;}v.push_back(p->data - '0');int temp = 0;for (int i = level; i>-1;i--){temp += v[i];/***从数组后面向前插入***/if (temp == sum){print(v,i);}}vector<int>vl(v),vr(v);findpath(p->left,sum, vl, level+1);findpath(p->right,sum, vr, level+1);}/***从下向上，打印的是从该节点向上i层的data***/void print(vector<int>v, int i){if (v.empty() || i<0){return;}for (int j = i;j < v.size(); j++){cout<<v[j]<<"  ";}cout<<endl;}public:tree(){//root = create();root = NULL;index = 0;}/***输入扩展层次遍历序列，#表示该节点为空***/tree(char *s){root = NULL;index = 0;if (s == NULL){return;}int size = strlen(s);create(&root,s,0,size,NULL);}~tree(){/***清空二叉树***/}void printPathSum(int sum){findpath(root, sum);}void printPathSum2(int sum){vector<int>v;findpath(root, sum,v,0);}void preOrder(){pOrder(root);}void inOreder(){zOrder(root);}void postOreder(){  hOrder(root);}};int main(){/***扩展层次序列简立树***/char t[14] = "1234#54#6##23";tree s(t);//s.preOrder();s.printPathSum(10);cout<<"xxxx"<<endl;s.printPathSum2(10);cout<<"Over"<<endl;return 0; }