LeetCode

Appearance

101. Symmetric Tree ​

题目 ​

Given the root of a binary tree, check whether it is a mirror of itself (i.e., symmetric around its center).

 

Example 1:

Input: root = [1,2,2,3,4,4,3]
Output: true

Example 2:

Input: root = [1,2,2,null,3,null,3]
Output: false

 

Constraints:

  • The number of nodes in the tree is in the range [1, 1000].
  • -100 <= Node.val <= 100

 

Follow up: Could you solve it both recursively and iteratively?
Related Topics
  • 树
  • 深度优先搜索
  • 广度优先搜索
  • 二叉树

  • 👍 2975
  • 👎 0

    • 递归思路 ​

    递归判断每个左右节点的数据是否相同，不同可以直接返回 False，终止递归判断

    解法 ​

    python
    class Solution:
    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
        def check(p: Optional[TreeNode], q:Optional[TreeNode]) -> bool:
            if not p and not q: return True
            if not p or not q: return False
            return p.val == q.val and check(p.left, q.right) and check(p.right, q.left)

        return check(root, root)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(N)

    迭代法 ​

    广度优先遍历，注意点进入对列的顺序为

    • leftTreeNode.left

    • rightTreeNode.right

    • leftTreeNode.right

    • rightTreeNode.left

    py
    # leetcode submit region begin(Prohibit modification and deletion)
# 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:
        def check(p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
            quene = []
            quene.append(p)
            quene.append(q)

            while quene:
                l = quene.pop(0)
                r = quene.pop(0)
                if not l and not r:
                    continue
                if not l or not r:
                    return False
                if l.val != r.val:
                    return False

                quene.append(l.left)
                quene.append(r.right)

                quene.append(l.right)
                quene.append(r.left)

            return True

        return check(root, root)

        
# leetcode submit region end(Prohibit modification and deletion)

    复杂度分析 ​

    • 时间复杂度 O(N)
    • 空间复杂度 O(N)