【二叉树】101.对称二叉树

题目

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
        if root == None or (root.left == None and root.right == None):
            return True
        return self.myJudge(root.left, root.right)
    
    def myJudge(self, left: Optional[TreeNode], right: Optional[TreeNode]):
        if left == None and right == None:
            return True
        elif left == None or right == None:
            return False
        elif left.val != right.val:
            return False
        else:
            return self.myJudge(left.left, right.right) and self.myJudge(left.right, right.left)

Java

法1:基本好法

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if (root == null || (root.left == null && root.right == null)) {
            return true;
        }
        return myJudge(root.left, root.right);
    }

    public boolean myJudge(TreeNode left, TreeNode right) {
        if (left == null && right == null) {
            return true;
        } else if (left == null || right == null) {
            return false;
        } else if (left.val != right.val) {
            return false;
        }
        return myJudge(left.left, right.right) && myJudge(left.right, right.left);
    }
}