检测一个二叉树是否是另一个二叉树的子树

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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 <iostream>#include <cstring>using namespace std;#define MAX 100 // Structure of a tree nodestruct Node{    char key;    struct Node *left, *right;}; // A utility function to create a new BST nodeNode *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 referencevoid 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 referencevoid 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 functionint 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;}

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