101. Symmetric Tree

Edited by Charles

Easy · Python · Binary Tree

Problem

Read the original problem on LeetCode.

The official page contains the problem statement, examples, and constraints.

Explanation, examples & other approaches

Core idea

Recursion:

ultimate checks postorder traversal

Original study notes follow, including intermediate snippets and earlier attempts.

Recursion:

1. ultimate checks
   1. left: None v.s. right: not None → False
   2. left: not None v.s. right: None → False
   3. left: None v.s. right: None → True
   4. left.val ≠ right.val → False
   5. The reason why d is checked after a-c is left and right may be null. If d is checked in front of a, there will be errors since left.val or right.val does not exist.
2. postorder traversal
   1. check children first, i.e. outer and inner. These two checks are actually recursions.
   2. check the parent. That is, the parents are equal only when the outer and the inner are True.
   3. For each individual node and its corresponding to-be-compared node, they are equal only when both the outer and the inner are True.

# 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:
        left, right = root.left, root.right
        return self.checkSymmetric(left, right)

    def checkSymmetric(self, left, right):
        if not left and right:
            return False
        if left and not right:
            return False
        if not left and not right:
            return True
        if left.val != right.val:
            return False

        # Check left/right
        outer = self.checkSymmetric(left.left, right.right)
        # Check right/left
        inner = self.checkSymmetric(left.right, right.left)
        # Check middle
        return outer and inner

Iteration:

1. The basic idea is still the same.
2. Pay attention to the children nodes. Since, in every iteration, the queue pops out two elements, left.left and right.right should be appended to the queue in order and in pair; left.right and right.left should be appended in order and in pair as well.
3. The solution uses queue. Similarly, stack also works in this question.

Study record

Field	Value
Attempts	1
Mastery	❚
The Latest Attempt	January 13, 2026

Python solution

# 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
from collections import deque

class Solution:
    def isSymmetric(self, root: Optional[TreeNode]) -> bool:
        dq = deque()
        dq.append(root.left)
        dq.append(root.right)

        while dq:
            left = dq.popleft()
            right = dq.popleft()

            if not left and not right:
                continue

            if not left and right:
                return False
            if left and not right:
                return False
            if left.val != right.val:
                return False

            dq.append(left.left)
            dq.append(right.right)
            dq.append(left.right)
            dq.append(right.left)

        return True

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