Sword finger offer 26 Substructure of tree

Original link ： The finger of the sword Offer 26. The substructure of a tree

solution:

  The code is mainly divided into two steps ：

        ① adopt dfs Judge B Yes no A The subtree of the root node
        ② adopt isSubstructure Function for each A Traverse the nodes in

class Solution {
public:
    bool isSubStructure(TreeNode* A, TreeNode* B) {
        if(B == NULL || A == NULL) return false;
        return dfs(A, B) || isSubStructure(A->left, B) || isSubStructure(A->right, B);
    }

    bool dfs(TreeNode *A, TreeNode *B) {
        if(B == NULL)
            return true;
        if(A == NULL || A->val != B->val)
            return false;
        return dfs(A->left, B->left) && dfs(A->right, B->right);
    }
};