請實現一個函數,用來判斷一顆二叉樹是不是對稱的。
注意,如果一個二叉樹同此二叉樹的鏡像是同樣的,定義其爲對稱的。
思路:針對前序遍歷定義一種對稱的前序遍歷算法:先遍歷父節點,再遍歷它的右子節點,最後遍歷它的左子節點
python:
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isSymmetrical(self, pRoot):
# write code here
return self.isSymmetrical_2(pRoot,pRoot)
def isSymmetrical_2(self,pRoot1,pRoot2):
if pRoot1==None and pRoot2==None:
return True
if pRoot1==None or pRoot2==None:
return False
if pRoot1.val != pRoot2.val:
return False
return self.isSymmetrical_2(pRoot1.left,pRoot2.right)
c++:
struct TreeNode {
int val;
struct TreeNode *left;
struct TreeNode *right;
TreeNode(int x) :
val(x), left(NULL), right(NULL) {
}
};
*/
class Solution {
public:
bool isSymmetrical(TreeNode* pRoot)
{
return isSymmetrical(pRoot, pRoot);
}
bool isSymmetrical(TreeNode* pRoot1, TreeNode* pRoot2)
{
if (pRoot1 == nullptr && pRoot2 == nullptr)
return true;
if (pRoot1 == nullptr || pRoot2 == nullptr)
return false;
if (pRoot1->val != pRoot2->val)
return false;
return isSymmetrical(pRoot1->left, pRoot2->right);
}
};