Symmetric Tree
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.
和same tree题目差不多一样的解法。
1 recursively 递归法:
//2014-2-15 update
bool isSymmetric(TreeNode *root)
{
if (!root) return true;
return isSymmetric(root->left, root->right);
}
bool isSymmetric(TreeNode *lt, TreeNode *rt)
{
if (!lt && !rt) return true;
if (lt && !rt || !lt && rt || lt->val != rt->val) return false;
return isSymmetric(lt->left, rt->right) &&isSymmetric(lt->right, rt->left);
}
2 iteratively 非递归:
使用先序遍历,原来树的先序遍历的非递归写法也不简单。
//2014-2-15 update
bool isSymmetric(TreeNode *root)
{
if (!root || !root->left && !root->right) return true;
TreeNode *t1 = root->left, *t2 = root->right;
if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;
stack<TreeNode *> s1, s2;
s1.push(t1), s2.push(t2);
bool flag = false;
while (!s1.empty() && !s2.empty())
{
if (!flag && (t1->left || t2->right))
{
s1.push(t1), s2.push(t2);
t1 = t1->left, t2 = t2->right;
if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;
s1.push(t1), s2.push(t2);
}
else if (t1->right || t2->left)
{
t1 = t1->right, t2 = t2->left;
if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;
flag = false;
}
else
{
t1 = s1.top(), t2 = s2.top();
s1.pop(), s2.pop();
flag = true;
}
}
return s1.empty() && s2.empty();
}
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