Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree [1,2,2,3,4,4,3]
is symmetric:
1
/ \
2 2
/ \ / \
3 4 4 3
But the following [1,2,2,null,3,null,3]
is not:
1
/ \
2 2
\ \
3 3
Follow up: Solve it both recursively and iteratively.
题意
判断是否是镜像二叉树
思路1
- 均为叶节点时满足
- 一个节点为叶节点另一个不是时不满足
- 非叶节点需要满足值相等,左子树和右子树相等,右子树和左子树相等
代码1
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode* root) {
return isMirror(root, root);
// 等价于
// if(root == NULL) return true;
// else return isMirror(root->left, root->right);
}
bool isMirror(TreeNode* t1, TreeNode* t2) {
if(t1 == NULL && t2 == NULL) return true; // 叶结点
if(t1 == NULL || t2 == NULL) return false; // 不符合
return t1->val==t2->val && isMirror(t1->left, t2->right) && isMirror(t1->right, t2->left);
}
};