Easy
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:
[1, 1000]
.-100 <= Node.val <= 100
Follow up: Could you solve it both recursively and iteratively?
import { TreeNode } from '../../com_github_leetcode/treenode'
/*
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function isSymmetric(root: TreeNode | null): boolean {
if (!root.left && !root.right) return true
const queue: [TreeNode, TreeNode][] = [[root.left, root.right]]
while (queue.length > 0) {
let qLen: number = queue.length
while (qLen-- > 0) {
const [leftNode, rightNode] = queue.shift()
if (!leftNode && !rightNode) continue
if (!leftNode || !rightNode || leftNode.val != rightNode.val) return false
queue.push([leftNode.left, rightNode.right])
queue.push([leftNode.right, rightNode.left])
}
}
return true
}
export { isSymmetric }