17. 树的子结构

/*
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 res = false;
        //任意一个为空，就返回false
        if(pRoot1 != NULL && pRoot2 != NULL){
        // 如果根节点相同，则判断是否是子树关系
            if(pRoot1->val == pRoot2->val){
                res = Tree1HasTree2(pRoot1, pRoot2);
            }
            //若不是子树，则进一步判断pRoot1的左子树和右子树是否包括pRoot2 
            if(!res){
                res = HasSubtree(pRoot1->left, pRoot2);
            }
            if(!res){
                res = HasSubtree(pRoot1->right, pRoot2);
            } 
        }       
        return res;   
    }
    bool Tree1HasTree2(TreeNode* node1, TreeNode* node2){
    // 一定要先判断子树是否已经遍历完
        if(node2 == NULL) 
            return true;
        if(node1 == NULL) 
            return false;
        if(node1->val != node2->val){
            return false;
        }
        return Tree1HasTree2(node1->left, node2->left) && Tree1HasTree2(node1->right, node2->right);
    }
};