Table of contents
topic
Symmetric binary tree judgment
Topic requirements
example
answer
method one,
recursive method
Implementation ideas
The method of judging whether two binary trees are equal is used, but the symmetric binary tree judges whether the left child of p node and the right child of q node, the right child of p node and the left child of q node are equal.
Time Complexity and Space Complexity
Time complexity: O(N)
Space complexity: O(N)
the code
bool isSameTree(struct TreeNode* p, struct TreeNode* q){
if(p==NULL&&q==NULL)
{
return true;
}
if(p==NULL||q==NULL)
{
return false;
}
if(p->val!=q->val)
{
return false;
}
//和判断两个二叉树相等类似，不过判断的是p结点的左孩子和q结点的右孩子是否相等
//还判断p结点的右孩子和q结点的左孩子是否相等，因为对称为镜像，所以需要这样比较。
return isSameTree(p->left,q->right)&&isSameTree(p->right,q->left);
}
bool isSymmetric(struct TreeNode* root){
//如果二叉树为空，则返回true
if(root==NULL)
{
return true;
}
return isSameTree(root->left,root->right);
}
Method Two,
Implementation ideas
Time Complexity and Space Complexity
the code