在C中搜索二叉树中的节点答案

【问题标题】：Search a node in a binary tree in C在C中搜索二叉树中的节点
【发布时间】：2021-07-12 03:18:15
【问题描述】：

我有一棵像这样的二叉树（不是二叉搜索树，BST）

// node in a tree
typedef struct
{
    int id;         // node index in a tree
    int value;      // data on node
    struct Node* l; // left child
    struct Node* r; // right child
} Node;

// a tree with only its root node
typedef struct
{
    Node* root;
} BiTree;

我想通过其id（对于给定索引idToBeFound）找到一个节点并返回找到的节点的指针。我的搜索功能findNode 适用于下面的case 1，但不适用于case 2（似乎进入了无限循环）：

示例（此处显示的数字是节点的索引和value=id 为简单起见）：如果查找 id = 2 的节点，则输出应该是节点 2 的指针。

case 1:

case 2:

// finds a node in a tree and returns its pointer
Node* findNode(Node *node, int idToBeFound)
{
    printf("traversing node %.2d with value = %d\n", node->id, node->value);

    if (node == NULL)
    {
        printf("Failed to find!");
        return NULL;
    }

    if (idToBeFound == node->id)
    {
        printf("returning node %d\n", node->id);
        return node;
    }

    Node* node1 = findNode(node->l, idToBeFound);
    if(node1 != NULL)
    {
        printf("returning the left child of node %d\n", node->id);
        return node1;
    }

    Node* node2 = findNode(node->r, idToBeFound);
    if(node2 != NULL)
    {
        printf("returning the right child of node %d\n", node->id);
        return node2;
    }
}

int main(void)
{
    BiTree tree;
    initTree(&tree);

    //tree.root = (Node*) malloc(sizeof(Node));
    //createTreePreOderRecur(tree.root);        // create a tree recursively
    
    printf("Demo: find a node via its id in a tree:\n");
    int IdToBeFound = 2;
    printf("Id of node to be found = %d\n", IdToBeFound);

    Node* nodeFound = (Node*) malloc(sizeof(Node));
    nodeFound = findNode(tree.root, IdToBeFound);

    printf("the found node is: id = %d, value = %d\n", nodeFound->id, nodeFound->value);

    //destoryTreeRecur(tree.root);

    return 0;
}

函数findNode有什么问题吗？（其他代码运行良好，此处忽略）。谢谢。

【问题讨论】：

如果node1 和node2 都是NULL 会发生什么？看起来函数 findNode() 最后缺少 return NULL。你的编译器应该已经警告你了。如果没有，请打开编译器警告并修复所有警告。
Node* nodeFound = (Node*) malloc(sizeof(Node)); nodeFound = findNode(tree.root, IdToBeFound);
@wildplasser：我释放了nodeFound，但问题仍然存在，所以可能不是由于内存泄漏。
@G。 Sliepen：我在if(node2 != NULL){...}后面加了一行代码return NULL;，用于你指出的情况，不起作用。
会不会是创建树的函数有bug？

标签： c search binary-tree

【解决方案1】：

我想我现在有正确的代码，可以正常运行：

Node* findNode(Node *node, int value)
{
    printf("traversing node %d with value = %d\n", node->id, node->value);

    Node *temp = NULL;

    // case 1:
    if (node->id == value)
    {
        return node;
    }

    // case 2: if the root node is not the right one, traverse its left child
    if (node->l != NULL)
    {
        printf("traversing the left child of node %d\n", node->id);
        temp = findNode(node->l, value);
        if (temp != NULL) return temp;
    }

    // case 3: if the left node is not the right one, traverse its right child
    if (node->r != NULL)
    {
        printf("traversing the right child of node %d\n", node->id);
        temp = findNode(node->r, value);
        if (temp != NULL) return temp;
    }

    // case 4: if the root node and its children are not the right one, return NULL
    return temp;
}

int main(void)
{
    BiTree tree;
    ...

    int IdToBeFound = 2;    
    printf("Demo: find a node via its id in a tree:\n");
    printf("The id of node to find is %d\n", IdToBeFound);

    Node* nodeFound = findNode(tree.root, IdToBeFound);
    
    /* 
    Check the result by evaluating the pointer nodeFound
    1: nodeFound = NULL means failure to find (e.g., the node is not in the tree), 
    2: nodeFound != NULL means success.
    */
    if(nodeFound != NULL) 
    {
        printf("the found node is: id = %d, value = %d\n", nodeFound->id, nodeFound->value);
    }
    else
    {
        printf("failed to find node %d in the tree!\n", IdToBeFound);
        free(nodeFound);
    }

    ...

    return 0;
}

【讨论】：