Binary Tree · DFS / BFS · LC 101

Symmetric Tree

Check whether a binary tree is a mirror of itself - left subtree mirrors the right subtree at every level.

Recursive DFSMirror CheckBFS VariantEasy

Key concepts at a glance — for those who already know the basics.

Problem Statement

Given the root of a binary tree, determine whether it is symmetric around its center.

Symmetric means left and right subtrees are mirror images with matching structure and matching values at mirrored positions.

Example: [1,2,2,3,4,4,3] => true, [1,2,2,null,3,null,3] => false

01 · The rule

A tree is symmetric if its left and right subtrees are mirrors of each other

Two subtrees are mirrors when: their roots have equal values, the left child of one equals the right child of the other, and vice versa - checked recursively all the way down.

02 · How to think

Start with the root's two children

Call a helper isMirror(left, right) with root.left and root.right.

Check the base cases first

Both None -> symmetric. One None, one not -> asymmetric. Both exist -> compare values.

Recurse with the outer and inner pairs

Outer: left.left vs right.right. Inner: left.right vs right.left. Both must pass.

Return True only when values match AND both recursive calls pass

Any mismatch anywhere in the subtree short-circuits the whole check to False.

03 · Code (Python)

def isSymmetric(root) -> bool:
 
    def isMirror(left, right) -> bool:
        if not left and not right:  # both None -> symmetric
            return True
        if not left or not right:   # one None -> not symmetric
            return False
 
        return (
            left.val == right.val          # values match
            and isMirror(left.left,  right.right)  # outer pair
            and isMirror(left.right, right.left)   # inner pair
        )
 
    return isMirror(root.left, root.right)

04 · Example tree & mirror check

Call

left

right

Result

isMirror(2, 2)

node(2)

2==2, recurse

isMirror(3, 3)
node(3)
node(3)
3==3, both children None => True

isMirror(4, 4)

node(4)

4==4, both children None => True

Final
-
-
True - tree is symmetric

05 · Complexity

Time

O(n)

Every node is visited exactly once in mirror comparisons.

Space

O(h)

Call stack depth equals tree height h.

Try it yourself

LeetCode 101 · Easy · Tags: Tree, DFS, BFS, Binary Tree

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