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剑指offer——面试题18：树的子结构

Solution1：

/*struct TreeNode {    int val;    struct TreeNode *left;    struct TreeNode *right;    TreeNode(int x) :            val(x), left(NULL), right(NULL) {    }};*/class Solution {public:    bool HasSubtree(TreeNode* pRoot1, TreeNode* pRoot2){        bool result = false;        //先找到相同的根结点的值        if(pRoot1 != NULL && pRoot2 != NULL){            if(pRoot1->val == pRoot2->val)                result = DoesTree1HaveTree2(pRoot1, pRoot2);            if(!result)                result = HasSubtree(pRoot1->left, pRoot2);            if(!result)                result = HasSubtree(pRoot1->right, pRoot2);        }        return result;    }    //判断子树结构是否一致    bool DoesTree1HaveTree2(TreeNode* pRoot1, TreeNode* pRoot2){        if(pRoot2 == NULL) //已经到达树2的根结点，返回true            return true;        if(pRoot1 == NULL) //已经到了树1的根结点，但还未到达树2的根结点，返回false            return false;        if(pRoot1->val != pRoot2->val) //值不相同，则返回false            return false;        //递归判断        return DoesTree1HaveTree2(pRoot1->left, pRoot2->left) && DoesTree1HaveTree2(pRoot1->right, pRoot2->right);    }};

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