Symmetric Tree
描述
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.
分析
无
递归版
// Symmetric Tree
// 递归版,时间复杂度O(n),空间复杂度O(logn)
class Solution {
public:
bool isSymmetric(TreeNode *root) {
if (root == nullptr) return true;
return isSymmetric(root->left, root->right);
}
bool isSymmetric(TreeNode *p, TreeNode *q) {
if (p == nullptr && q == nullptr) return true; // 终止条件
if (p == nullptr || q == nullptr) return false; // 终止条件
return p->val == q->val // 三方合并
&& isSymmetric(p->left, q->right)
&& isSymmetric(p->right, q->left);
}
};
迭代版
// Symmetric Tree
// 迭代版,时间复杂度O(n),空间复杂度O(logn)
class Solution {
public:
bool isSymmetric (TreeNode* root) {
if (!root) return true;
stack<TreeNode*> s;
s.push(root->left);
s.push(root->right);
while (!s.empty ()) {
auto p = s.top (); s.pop();
auto q = s.top (); s.pop();
if (!p && !q) continue;
if (!p || !q) return false;
if (p->val != q->val) return false;
s.push(p->left);
s.push(q->right);
s.push(p->right);
s.push(q->left);
}
return true;
}
};