二叉树--OJ2

1.判断两棵树是否相同

    bool sametree(TreeNode* root1,TreeNode* root2)
    {
        if(root1==NULL&&root2==NULL)
        return 1;
        if(root1==NULL||root2==NULL)
        return 0;
        if(root1->val!=root2->val)
        return 0;
        return sametree(root1->left,root2->left)&&sametree(root1->right,root2->right)
    }

2.另一棵树的子树

class Solution {
public:
    TreeNode* PreOrder(TreeNode* root,int a) {
        if (root == NULL) {
            return NULL;
        }
        if(root->val==a)
        {
            return root;
        }
        auto r1=PreOrder(root->left,a);
        if(r1)
        return r1;
        auto r2=PreOrder(root->right,a);
        return r2;
    }
    bool sametree(TreeNode* root1, TreeNode* root2) {
        if (root1 == NULL && root2 == NULL)
            return 1;
        if (root1 == NULL || root2 == NULL)
            return 0;
        if(root1->val!=root2->val)
        return 0;
        return sametree(root1->left, root2->left) &&
               sametree(root1->right, root2->right);
    }
    bool isSubtree(TreeNode* root, TreeNode* subRoot) {
        int a=subRoot->val;
        while(auto b=PreOrder(root, a))
        {
            if(sametree(b, subRoot))
            {
                return 1;
            }
            b->val=(!a);

        }
        return 0;
    }
};

法二

class Solution {
public:
    bool sametree(TreeNode* root1, TreeNode* root2) {
        if (root1 == NULL && root2 == NULL)
            return 1;
        if (root1 == NULL || root2 == NULL)
            return 0;
        if(root1->val!=root2->val)
        return 0;
        return sametree(root1->left, root2->left) &&
               sametree(root1->right, root2->right);
    }
    bool isSubtree(TreeNode* root, TreeNode* subRoot) {
        if(root==NULL)
        return 0;
        if(sametree(root, subRoot))
        return 1;
        return isSubtree(root->left, subRoot)||isSubtree(root->right, subRoot);
    }
};

对称二叉树

class Solution {
public:
    bool sametree(TreeNode* root1, TreeNode* root2) {
        if (root1 == NULL && root2 == NULL)
            return 1;
        if (root1 == NULL || root2 == NULL)
            return 0;
        if(root1->val!=root2->val)
        return 0;
        return sametree(root1->left, root2->left) &&
               sametree(root1->right, root2->right);
    }
    bool isSubtree(TreeNode* root, TreeNode* subRoot) {
        if(root==NULL)
        return 0;
        if(sametree(root, subRoot))
        return 1;
        return isSubtree(root->left, subRoot)||isSubtree(root->right, subRoot);
    }
};