在二元树中找出和为某一值的所有路径

最新推荐文章于 2017-08-04 21:42:41 发布

ruoshui13

最新推荐文章于 2017-08-04 21:42:41 发布

阅读量460

点赞数

CC 4.0 BY-SA版权

分类专栏：算法

本文链接：https://blog.youkuaiyun.com/yuanwei1314/article/details/41625243

算法专栏收录该内容

45 篇文章

订阅专栏

本文介绍了一种通过递归求解的方法，在二元树中查找与给定整数相等的所有路径。通过创建、遍历二元树并使用深度优先搜索策略，最终打印出所有满足条件的路径。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

题目：

输入一个整数和一棵二元树，从树的根结点开始往下访问一直到叶结点所经过的所有结点形成一条路径，然后打印出和与输入整数相等的所有路径。
例如输入整数22和如下二元树

    10
/   \
5    12
/ \
4 7

则打印出两条路径：10, 12和10, 5, 7。
其中，二元树节点的数据结构定义为：
```
struct BinaryTreeNode // a node in the binary tree
{
    int m_nValue; // value of node
    BinaryTreeNode *m_pLeft; // left child of node
    BinaryTreeNode *m_pRight; // right child of node
};
```

思路：递归求解

代码：

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct BinaryTreeNode // a node in the binary tree
{
    int m_nValue; // value of node
    BinaryTreeNode *m_pLeft; // left child of node
    BinaryTreeNode *m_pRight; // right child of node
};
/*创建二叉树*/
void creatTree(BinaryTreeNode* &bTree, int *a, int pos, int len)
{
     if (pos >= len)
     {
         bTree = NULL;
         return;
     }
     
     bTree = (BinaryTreeNode*)malloc(sizeof(BinaryTreeNode));
     bTree->m_pLeft = NULL;
     bTree->m_pRight = NULL;
     bTree->m_nValue = a[pos];
     creatTree(bTree->m_pLeft, a, 2*pos+1, len);
     creatTree(bTree->m_pRight, a, 2*pos+2, len);
}
/*先序遍历*/
void preOrderTree(BinaryTreeNode* bTree)
{
    if (bTree == NULL)
    {
        return;
    }
    printf("%d ", bTree->m_nValue);
    preOrderTree(bTree->m_pLeft);
    preOrderTree(bTree->m_pRight);
}
/*中序遍历*/
void inOrderTree(BinaryTreeNode* bTree)
{
    if (bTree == NULL)
    {
        return;
    }
    inOrderTree(bTree->m_pLeft);
    printf("%d ", bTree->m_nValue);
    inOrderTree(bTree->m_pRight);
}
/*后序遍历*/
void postOrderTree(BinaryTreeNode* bTree)
{
    if (bTree == NULL)
    {
        return;
    }
    postOrderTree(bTree->m_pLeft);
    postOrderTree(bTree->m_pRight);
    printf("%d ", bTree->m_nValue);
}

void printPath(int data[], int index)
{
    int i;
    for (i=0; i<index; i++)
    {
        printf("%d\t", data[i]);
    }
    printf("\n");
}
int tag = 22;
/*说明:
  sum为目标数当前和
  data为存储的结果
  index为当前存储数组的长度
*/
void findPath(BinaryTreeNode* bTree, int sum, int data[], int index)
{
   data[index++] = bTree->m_nValue;
    if (bTree->m_pLeft==NULL && bTree->m_pRight==NULL)
    {
        if (bTree->m_nValue+sum == tag)
        {
             printPath(data, index);
        }
    }
    /*找到左支*/
    if (bTree->m_pLeft != NULL)
    {
        findPath(bTree->m_pLeft, sum+bTree->m_nValue, data, index);
    }
    /*找到右支*/
    if (bTree->m_pRight != NULL)
    {
        findPath(bTree->m_pRight, sum+bTree->m_nValue, data, index);
    }
}

int main()
{
    int a[] = {10, 5, 12, 4, 7};
    BinaryTreeNode* bTree = NULL;
    int path[100];
    creatTree(bTree, a, 0, 5);
    printf("先序遍历二叉树:\n");
    preOrderTree(bTree);
    printf("\n");
    printf("中序遍历二叉树:\n");
    inOrderTree(bTree);
    printf("\n");
    printf("后序遍历二叉树:\n");
    postOrderTree(bTree);
    printf("\n");
    findPath(bTree, 0, path, 0);
    return 0;
}