LeetCode101 Symmetric Tree 对称树

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.

 题源：here；完整实现：here

思路：

这道题关键在于理解对称：左边等于右边，这还只理解到了一层，再深入一层：左边的左边等于右边的右边，左边的右边等于右边的左边，这两层理解了就知道怎么迭代和递归了。

1 递归

bool helper(TreeNode *left, TreeNode *right){
	if (!left && !right) return true;
	if (!left || !right) return false;
	return left->val == right->val
		&& helper(left->left, right->right)
		&& helper(left->right, right->left);
}

bool isSymmetric(TreeNode* root) {
	return helper(root, root);
}

2 迭代

迭代的进行需要使用到队列：先入先出。

bool isSymmetric2(TreeNode* root){
	queue<TreeNode* > records;
	records.push(root);
	records.push(root);
	while (records.size()){
		TreeNode *left = records.front(); records.pop();
		TreeNode *right = records.front(); records.pop();
		if (!left && !right) continue;
		else if (!left || !right) return false;
		else if (left->val != right->val) return false;
		records.push(left->left);
		records.push(right->right);
		records.push(left->right);
		records.push(right->left);
	}

	return true;
}