目录
题目1.226.翻转二叉树
题目2.101. 对称二叉树
题目1.226.翻转二叉树
力扣题目链接(opens new window)
听说一位巨佬面Google被拒了，因为没写出翻转二叉树 | LeetCode：226.翻转二叉树_哔哩哔哩_bilibili
翻转一棵二叉树。
Python
# Definition for a binary tree node.
# class TreeNode(object):
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution(object):def invertTree(self, root):""":type root: TreeNode:rtype: TreeNode"""if not root:return Noneroot.left, root.right = root.right, root.leftself.invertTree(root.left)self.invertTree(root.right)return rootC
/*** Definition for a binary tree node.* struct TreeNode {* int val;* struct TreeNode *left;* struct TreeNode *right;* };*/
struct TreeNode* invertTree(struct TreeNode* root) {if (root == NULL){return root;}struct TreeNode* temp = root->right;root->right = root->left;root->left = temp;invertTree(root->left);invertTree(root->right);return root;
}题目2.101. 对称二叉树
力扣题目链接(opens new window)
新学期要从学习二叉树开始！ | LeetCode：101. 对称二叉树_哔哩哔哩_bilibili
相似题目：
- 100.相同的树(opens new window)
- 572.另一个树的子树
给定一个二叉树，检查它是否是镜像对称的。
Python
class Solution:def isSymmetric(self, root: TreeNode) -> bool:if not root:return Truereturn self.compare(root.left, root.right)def compare(self, left, right):#首先排除空节点的情况if left == None and right != None: return Falseelif left != None and right == None: return Falseelif left == None and right == None: return True#排除了空节点，再排除数值不相同的情况elif left.val != right.val: return False#此时就是：左右节点都不为空，且数值相同的情况#此时才做递归，做下一层的判断outside = self.compare(left.left, right.right) #左子树：左、 右子树：右inside = self.compare(left.right, right.left) #左子树：右、 右子树：左isSame = outside and inside #左子树：中、 右子树：中 （逻辑处理）return isSameC
public bool IsSymmetric(TreeNode root)
{if (root == null) return true;return Compare(root.left, root.right);
}
public bool Compare(TreeNode left, TreeNode right)
{if(left == null && right != null) return false;else if(left != null && right == null ) return false;else if(left == null && right == null) return true;else if(left.val != right.val) return false;var outside = Compare(left.left, right.right);var inside = Compare(left.right, right.left);return outside&&inside;
}总结：解法只涉及到递归法，进阶的解法可以参考迭代法，难点在于理解递归的三步骤，考虑遍历的方式，是前中后遍历的哪一种，使用层序遍历等。
未完待续~
