Q: Input two binary trees A and B, and judge whether B is a substructure of A. (ps: we agree that an empty tree is not a substructure of any tree)
A: To find out whether there is a subtree in tree A with the same structure as tree B. We can divide it into two steps:
The first step is to find a node R with the same value as the root node of B in tree A,
The second step is to determine whether the subtree with R as the root node in tree A contains the same structure as tree B.
/*
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;
if (pRoot1 != NULL && pRoot2 != NULL)
{
if (pRoot1->val == pRoot2->val)
{
//After finding the same node, determine whether it contains the same structure
result = isSubTree(pRoot1, pRoot2);
}
if (!result)//Find the same node in the left subtree
{
result = HasSubtree(pRoot1->left, pRoot2);
}
if (!result)//Find the same node in the right subtree
{
result = HasSubtree(pRoot1->right, pRoot2);
}
}
return result;
}
bool isSubTree(TreeNode* pRoot1, TreeNode* pRoot2)
{
if(NULL == pRoot2)
{
return true;
}
if(NULL == pRoot1)
{
return false;
}
if(pRoot1->val != pRoot2->val)
{
return false;
}
return isSubTree(pRoot1->left, pRoot2->left) && isSubTree(pRoot1->right, pRoot2->right);
}
};