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：