leetcode || 101、Symmetric Tree

problem：

Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:

1
/ \
2   2
/ \ / \
3  4 4  3

But the following is not:

1
/ \
2   2
\   \
3    3

Note:

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

"{1,#,2,3}"

means? >
read more on how binary tree is serialized on OJ.

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题意：判断一棵二叉树是否为镜像二叉树（结构对称，数字相等）

thinking：

（1）该题合适的解法是递归法，采用DFS思想

递归，保存左右两个节点，然后判断leftNode->left和rightNode->right，以及leftNode->right和rightNode->left。如此不断递归

（2）还想到另外一种非递归法：层序遍历法，子结点为NULL时代表数值0，从左往右层序遍历得到数组a,从右往左层序遍历得到数组b

如果a==b，则该二叉树是镜像的。但提交时显示：Status:

Memory Limit Exceeded

code：

递归法：

class Solution {
public:
bool check(TreeNode *leftNode, TreeNode *rightNode)
{
if (leftNode == NULL && rightNode == NULL)
return true;

if (leftNode == NULL || rightNode == NULL)
return false;

return leftNode->val == rightNode->val && check(leftNode->left, rightNode->right) &&
check(leftNode->right, rightNode->left);
}

bool isSymmetric(TreeNode *root) {
if (root == NULL)
return true;

return check(root->left, root->right);
}
};

层序遍历法：Status:

Memory Limit Exceeded

class Solution {
private:
vector<int> res1;
vector<int> res2;
public:
bool isSymmetric(TreeNode *root) {
if(root==NULL)
return true;
res1.clear();
res2.clear();
level_walk_left(root);
level_walk_right(root);
if(res1==res2)
return true;
else
return false;
}
protected:
void level_walk_left(TreeNode *root)
{
queue<TreeNode *> _queue;
_queue.push(root);
while(!_queue.empty())
{
TreeNode *tmp=_queue.front();
if(tmp==NULL)
res1.push_back(0);
else
{
res1.push_back(tmp->val);
_queue.pop();
if(tmp->left!=NULL)
_queue.push(tmp->left);
else
_queue.push(NULL);
if(tmp->right!=NULL)
_queue.push(tmp->right);
else
_queue.push(NULL);
}
}
}

void level_walk_right(TreeNode *root)
{
queue<TreeNode *> _queue;
_queue.push(root);
while(!_queue.empty())
{
TreeNode *tmp=_queue.front();
if(tmp==NULL)
res2.push_back(0);
else
{
res2.push_back(tmp->val);
_queue.pop();
if(tmp->right!=NULL)
_queue.push(tmp->right);
else
_queue.push(NULL);
if(tmp->left!=NULL)
_queue.push(tmp->left);
else
_queue.push(NULL);
}
}
}

};