101 Symmetric Tree 对称二叉树
Description:
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
Example:
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.
题目描述:
给定一个二叉树，检查它是否是镜像对称的。
示例:
例如，二叉树 [1,2,2,3,4,4,3] 是对称的。
1
/ \
2 2
/ \ / \
3 4 4 3
但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:
1
/ \
2 2
\ \
3 3
说明:
如果你可以运用递归和迭代两种方法解决这个问题，会很加分。
思路:
本题与LeetCode #100 Same Tree 相同的树思路类似
不过是将根结点的左子树和根结点的右子树对称比较, 可以用递归(C++/Python)/迭代(Java)方式
时间复杂度O(n), 空间复杂度O(n), n为树中结点数量
代码:
C++:
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution
{
public:
bool isSymmetric(TreeNode* root)
{
if (!root) return true;
return isMirror(root, root);
}
private:
bool isMirror(TreeNode* p, TreeNode* q)
{
if (!p and !q) return true;
if (!p or !q) return false;
if (p -> val != q -> val) return false;
return isMirror(p -> left, q -> right) and isMirror(p -> right, q -> left);
}
};
Java:
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
public boolean isSymmetric(TreeNode root) {
if (root == null) return true;
Stack<TreeNode> s1 = new Stack<>();
Stack<TreeNode> s2 = new Stack<>();
s1.push(root);
s2.push(root);
while (!s1.isEmpty() && !s2.isEmpty()) {
TreeNode left = s1.pop();
TreeNode right = s2.pop();
if (left == null && right == null) continue;
if (left == null || right == null) return false;
if (left.val != right.val) return false;
s1.push(left.left);
s1.push(left.right);
s2.push(right.right);
s2.push(right.left);
}
return s1.isEmpty() && s2.isEmpty();
}
}
Python:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def isSymmetric(self, root: TreeNode) -> bool:
if not root:
return True
return self.isMirror(root, root)
def isMirror(self, p: TreeNode, q: TreeNode) -> bool:
if not p and not q:
return True
elif not p or not q:
return False
elif p.val != q.val:
return False
else:
return self.isMirror(p.left, q.right) and self.isMirror(p.right, q.left)
