检测一个二叉树是否是另一个二叉树的子树

Given two binary trees, check if the first tree is subtree of the second one. A subtree of a tree T is a tree S consisting of a node in T and all of its descendants in T.

The subtree corresponding to the root node is the entire tree; the subtree corresponding to any other node is called a proper subtree.

For example, in the following case, Tree1 is a subtree of Tree2.

给定两个二叉树，检测一个是否另一个的子树。例如下面的例子，Tree1是Tree2的子树：

        Tree1
          x  
        /    \ 
      a       b
       \
        c


        Tree2
              z
            /   \
          x      e
        /    \     \
      a       b      k
       \
        c

这里讨论的是O(N)的方案。这个想法是基于前序、中序（后序）可以唯一标识一个二叉树。如果树S前序、中序遍历后存储都是树T遍历存储后的字串，那么就是子树啦。

答案也比较直接：

#include 
#include 
using namespace std;
#define MAX 100
 
// Structure of a tree node
struct Node
{
    char key;
    struct Node *left, *right;
};
 
// A utility function to create a new BST node
Node *newNode(char item)
{
    Node *temp =  new Node;
    temp->key = item;
    temp->left = temp->right = NULL;
    return temp;
}
 
// A utility function to store inorder traversal of tree rooted
// with root in an array arr[]. Note that i is passed as reference
void storeInorder(Node *root, char arr[], int &i)
{
    if (root == NULL) return;
    storeInorder(root->left, arr, i);
    arr[i++] = root->key;
    storeInorder(root->right, arr, i);
}
 
// A utility function to store preorder traversal of tree rooted
// with root in an array arr[]. Note that i is passed as reference
void storePreOrder(Node *root, char arr[], int &i)
{
    if (root == NULL) return;
    arr[i++] = root->key;
    storePreOrder(root->left, arr, i);
    storePreOrder(root->right, arr, i);
}
 
/* This function returns true if S is a subtree of T, otherwise false */
bool isSubtree(Node *T, Node *S)
{
    /* base cases */
    if (S == NULL)  return true;
    if (T == NULL)  return false;
 
    // Store Inorder traversals of T and S in inT[0..m-1]
    // and inS[0..n-1] respectively
    int m = 0, n = 0;
    char inT[MAX], inS[MAX];
    storeInorder(T, inT, m);
    storeInorder(S, inS, n);
    inT[m] = '\0', inS[n] = '\0';
 
    // If inS[] is not a substring of preS[], return false
    if (strstr(inT, inS) == NULL)
        return false;
 
    // Store Preorder traversals of T and S in inT[0..m-1]
    // and inS[0..n-1] respectively
    m = 0, n = 0;
    char preT[MAX], preS[MAX];
    storePreOrder(T, preT, m);
    storePreOrder(S, preS, n);
    preT[m] = '\0', preS[n] = '\0';
 
    // If inS[] is not a substring of preS[], return false
    // Else return true
    return (strstr(preT, preS) != NULL);
}
 
// Driver program to test above function
int main()
{
    Node *T = newNode('z');
    T->left = newNode('x');
    T->right = newNode('e');
    T->left->left = newNode('a');
    T->left->right = newNode('b');
    T->right->right = newNode('k');
    T->left->left->right = newNode('c');
 
    Node *S = newNode('x');
    S->left = newNode('a');
    S->right = newNode('b');
    S->left->right = newNode('c');
 
    if (isSubtree(T, S))
        cout << "Yes: S is a subtree of T";
    else
        cout << "No: S is NOT a subtree of T";
 
    return 0;
}