求二叉树中和为给定值的所有路径

转自：http://blog.csdn.net/lalor/article/details/7614381

问题定义：

        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.

 

解题思路：

      一层一层的遍历，保存当前节点到根节点的完整路径，然后从当前节点向上扫描，如果找到了当前节点到某个节点的和等于给定值，则输出之。程序对每个节点都需要遍历一遍，还要扫描当前节点到根节点的路径，且需要保存每个节点到根节点的路径，所以时间复杂度为O（nlgn)，空间复杂度为O(nlgn)。（ps：关于本程序中建树部分，可以参考：http://blog.csdn.net/lalor/article/details/7613621）

 

代码实例：

#include <algorithm>  
#include <iostream>  
#include <time.h>  
#include <assert.h>  
#include <stdio.h>  
#include <vector>  
  
  
using namespace std;  
  
struct node   
{  
    int data;  
    struct node * lchild;  
    struct node * rchild;  
};  
  
  
//将数组转换为深度最低的二叉树，采用了二分查找的思想  
struct node* ConvertArrayToTree(int data[], int first, int last)  
{  
    if (last < first)   
    {  
        return NULL;  
    }  
    else  
    {  
        int mid = ( last + first ) / 2;  
        struct node * newNode = NULL;  
        newNode = (struct node *)malloc(sizeof(struct node));  
        newNode->data = data[mid];  
        newNode->lchild = ConvertArrayToTree(data, first, mid - 1);  
        newNode->rchild = ConvertArrayToTree(data, mid + 1, last);  
        return newNode;  
    }  
}  
  
//再最左边插入一个节点  
void InsertNodeAtLeft(struct node *root, struct node *newNode)  
{  
    assert(root != NULL && newNode != NULL);  
    while(root->lchild != NULL)  
    {  
        root = root->lchild;  
    }  
    root->lchild = newNode;  
}  
  
//在最右边插入一个节点  
void InsertNodeAtRight(struct node *root, struct node *newNode)  
{  
    assert(root != NULL && newNode != NULL);  
    while(root->rchild != NULL)  
    {  
        root = root->rchild;  
    }  
    root->rchild = newNode;  
}  
//中序遍历  
void Traverse(struct node *root)  
{  
    if (root == NULL)   
    {  
        return;  
    }  
    Traverse(root->lchild);  
    Traverse(root->rchild);  
    printf("%d\t", root->data);  
}  
  
//打印和为sum的路径  
void print(vector<int>& buffer, int first, int last)  
{  
    int i;  
    for (i = first; i <= last; i++)   
    {  
        cout << buffer[i] << "\t";  
    }  
    cout << endl;  
}  
void findSum(struct node *head, int sum, vector<int> &buffer, int level)  
{  
    if (head == NULL) return;  
  
    int i;  
    int tmp = sum;  
    buffer.push_back(head->data);  
    for (i = level; i >= 0; i--)  
    {  
        tmp -= buffer[i];  
        if (tmp == 0) print(buffer, i, level);  
    }  
  
    vector<int> lbuffer(buffer);  
    vector<int> rbuffer(buffer);  
  
    findSum(head->lchild, sum, lbuffer, level + 1);  
    findSum(head->rchild, sum, rbuffer, level + 1);  
}  
  
int main(int argc, char* argv[])  
{  
    const int SIZE = 10;//测试的数据量  
    int data[SIZE];//保存数据  
    int i, j;  
    struct node *head = NULL;  
  
    for (i = 0; i < SIZE; i++)   
    {  
        data[i] = i + 1;  
    }  
  
    head = ConvertArrayToTree(data, 0, SIZE - 1);  
  
    struct node *one = (struct node *)malloc(sizeof(struct node));  
    struct node *two = (struct node *)malloc(sizeof(struct node));  
    one->data = 11;  
    one->lchild = NULL;  
    one->rchild = NULL;  
      
    two->data = 4;  
    two->lchild = NULL;  
    two->rchild = NULL;  
  
    InsertNodeAtLeft(head, one);  
    InsertNodeAtRight(head, two);  
    //遍历数据  
//  Traverse(head);  
//  printf("\n");  
    vector<int> v;  
    findSum(head, 14, v, 0);  
    return 0;  
} 

该示例中所使用的二叉树如下所示：

 

运行结果如下：