101. 对称二叉树

题目来源：101. 对称二叉树

给定一个二叉树，检查它是否是镜像对称的。

 

例如，二叉树 [1,2,2,3,4,4,3] 是对称的。

    1
   / \
  2   2
 / \ / \
3  4 4  3

 

但是下面这个 [1,2,2,null,3,null,3] 则不是镜像对称的:

    1
   / \
  2   2
   \   \
   3    3

 

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {boolean}
 */
var isSymmetric = function(root) {
  var check = (left, right) => {
    if(!left && !right){
      return true
    }
    if(!left || !right){
      return false
    }
    return left.val === right.val && check(left.left, right.right) && check(left.right, right.left)
  }
  if(!root){
    return true
  }
  return check(root, root)
};

var isSymmetric = function(root) {
  var check = (left, right) => {
    let stackL = []
    let stackR = []
    stackL.push(left)
    stackR.push(right)

    while(stackL.length){
      let l = stackL.shift()
      let r = stackR.shift()
      if(!l && !r){
        continue
      }
      if((!l || !r) || (l.val !== r.val)){
        return false
      }

      stackL.push(l.left)
      stackL.push(l.right)
  
      stackR.push(r.right)
      stackR.push(r.left)
    }
    return true
  }
  return check(root, root)
};

Python3

# Definition for a binary tree node.
from _typeshed import SupportsReadline


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: TreeNode) -> bool:
      def check(l, r):
        if not l and not r:
          return True
        if not l or not r:
          return False
        return (l.val == r.val) and check(l.left, r.right) and check(l.right, r.left)

      return check(root, root)

      
class Solution:
    def isSymmetric(self, root: TreeNode) -> bool:
      def check(l, r):
        stackL = list()
        stackR = list()

        stackL.append(l)
        stackR.append(r)
        while stackL:
          left = stackL.pop(0)
          right = stackR.pop(0)
          if not left and not right:
            continue
          if (not left or not right) or (left.val != right.val):
            return False

          stackL.append(left.left)
          stackL.append(left.right)

          stackR.append(right.right)
          stackR.append(right.left)
        return True

      return check(root, root)

 

进阶：

你可以运用递归和迭代两种方法解决这个问题吗？