binary-tree - 如何高效地找到二叉树中给定节点(或项)的镜像节点

Question

一直在思考这个问题，一直没有找到好的、高效的解决方案。

如何在二叉树中找到给定节点(或项)的镜像节点？

// Node definition
struct _Node {
    char data;
    struct _Node* left;
    struct _Node* right;
} Node;

// Assumption:
//     "given" is guaranteed in the binary tree ("root" which is not NULL)
Node* FindMirrorNode(Node* root, Node* given)
{
     // Implementation here
}

// OR: 

// Assumption:
//    in the binary tree ("root"), there is no repeated items, which mean in each node the char data is unique;
//   The char "given" is guaranteed in the binary tree.
char FindMirrorNodeData(Node* root, char given)
{
    // Implementation here
}

注意:我不是在问如何找到给定树的镜像树:-)

For example, considering the tree below
              A
          /      \
       B             C
      /            /   \
    D             E     F
     \           / \
      G         H   I

The mirror node of 'D' is node 'F'; while the mirror node of 'G' is NULL.

谢谢。

Answer 1

我已经用 char 为函数编写了一个解决方案。是 FindMirrorNode(r, n) == FindMirrorNodeData(r, n->data) 吗？

您必须遍历整个树以搜索给定数据，同时将镜像节点保留在堆栈中。这是一个非常简单的解决方案，仍然非常有效。如果您愿意，可以将尾调用转换为 while。

static Node* FindMirrorNodeRec(char given, Node* left, Node* right)
{
    // if either node is NULL then there is no mirror node
    if (left == NULL || right == NULL)
        return NULL;
    // check the current candidates
    if (given == left->data)
        return right;
    if (given == right->data)
        return left;
    // try recursively
    // (first external then internal nodes)
    Node* res = FindMirrorNodeRec(given, left->left, right->right);
    if (res != NULL)
        return res;
    return FindMirrorNodeRec(given, left->right, right->left);
}

Node* FindMirrorNodeData(Node* root, char given)
{
    if (root == NULL)
        return NULL;
    if (given == root->data)
        return root;
    // call the search function
    return FindMirrorNodeRec(given, root->left, root->right);
}