Prove safety offer 18: sub-tree

Title Description

Two binary inputs A, B, B is judged not substructure A's. (Ps: we agreed empty tree is not a tree any substructure)

Problem-solving ideas

Verify B is not A subtree, intuitive approach A traversing the tree in any order, once the child node and the root node B are the same, this will be the subtree rooted at the node B tree and comparison is satisfied lookup successful, otherwise lookup fails. Tree preorder traversal of the most intuitive, here to preorder, for example, gives C ++ implementation code:

/ * 
Struct the TreeNode { 
    int Val; 
    struct the TreeNode * left; 
    struct the TreeNode * right; 
    the TreeNode (int X): 
            Val (X), left (NULL), right (NULL) { 
    } 
}; * / 
class Solution {
 public :
     BOOL HasSubtree (* pRoot1 the TreeNode, the TreeNode * pRoot2) 
    {        BOOL isSub = false ;
            // appears a two arbitrary tree is empty, return false 
           IF (pRoot1 pRoot2 = = NULL &&!! NULL) {
                IF (pRoot1-> Val pRoot2- ==> Val) { 
                   isSub =doesTree1HaveTree2(pRoot1,pRoot2);
               }if(!isSub){
                   isSub=HasSubtree(pRoot1->left,pRoot2);
               }if(!isSub){
                  isSub=HasSubtree(pRoot1->right,pRoot2); 
               }
           }
         return isSub;
    }
    bool doesTree1HaveTree2(TreeNode* pRoot1, TreeNode* pRoot2){
        if(pRoot2==NULL){
            return true;
        }
        if(pRoot1==NULL){
            return false;
        }
        if(pRoot1->val!=pRoot2->val){
            return false;
        }
        return doesTree1HaveTree2(pRoot1->left,pRoot2->left) && doesTree1HaveTree2(pRoot1->right,pRoot2->right);
    }
};