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给定一个二叉树,检查它是否是镜像对称的。
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
进阶:
你可以运用递归和迭代两种方法解决这个问题吗?
# 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 root is None:
return True
return self.is_symmetric(root.left, root.right)
def is_symmetric(self, left: TreeNode, right: TreeNode) -> bool:
if left is None and right is None:
return True
if left is None or right is None or left.val != right.val:
return False
return self.is_symmetric(left.left, right.right) and self.is_symmetric(left.right, right.left)
/**
* 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;
return isSymmetric(root.left, root.right);
}
private boolean isSymmetric(TreeNode left, TreeNode right) {
if (left == null && right == null) return true;
if (left == null || right == null || left.val != right.val) return false;
return isSymmetric(left.left, right.right) && isSymmetric(left.right, right.left);
}
}
class Solution {
public:
bool isSymmetric(TreeNode* root) {
if (!root) return true;
return isSymmetric(root->left, root->right);
}
private:
bool isSymmetric(TreeNode* left, TreeNode* right) {
if (!left && !right) return true;
if (!left && right || left && !right || left->val != right->val) return false;
return isSymmetric(left->left, right->right) && isSymmetric(left->right, right->left);
}
};
/**
* Definition for a binary tree node.
* function TreeNode(val) {
* this.val = val;
* this.left = this.right = null;
* }
*/
/**
* @param {TreeNode} root
* @return {number[][]}
*/
var levelOrder = function (root) {
let result = [];
if (!root) return result;
let queue = [];
queue.push(root);
while (queue.length) {
let size = queue.length;
let levelItems = [];
while (size--) {
let node = queue.shift();
levelItems.push(node.val);
if (node.left) {
queue.push(node.left);
}
if (node.right) {
queue.push(node.right);
}
}
result.push(levelItems);
}
return result;
};
### **...**
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