017-101-Symmetric Tree 判断树是否对称

Question

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

Note:

Bonus points if you could solve it both recursively and iteratively.

Solution

通过观察，树的镜像是指每个节点的左右子树互换，先遍历右子树，并交换每个节点的左右子树，然后判断左右子树是否相等参考 007-100-判断两个二叉树是否相等 Same Tree

/**
* Definition for a binary tree node.
* struct TreeNode {
*     int val;
*     struct TreeNode *left;
*     struct TreeNode *right;
* };
*/
void switchNode(struct TreeNode* root){
if(root != NULL){
struct TreeNode* tmp = root->left;
root->left = root->right;
root->right = tmp;

switchNode(root->left);
switchNode(root->right);
}
}

bool isSameTree(struct TreeNode* p, struct TreeNode* q) {
if(p == NULL || q == NULL){
return (p == q);
}else{
if(p->val != q->val){
return false;
}else{
return isSameTree(p->left, q->left)&&isSameTree(p->right, q->right);
}
}
}

bool isSymmetric(struct TreeNode* root) {

if(root != NULL){
switchNode(root->right);
return isSameTree(root->left, root->right);
}
return true;
}

时间效率是O(n), 这个解法唯一不好的就是改变了输入的结构。