牛客题霸-判读二叉树是否对称-Java题解

题目：https://www.nowcoder.com/practice/1b0b7f371eae4204bc4a7570c84c2de1

• 题解：这道题需要注意的地方就是对称的意思的理解，当root的左右孩子left和right的数值相等，同时left.left和right.right以及left.right和right.left分别对应对称时，root是对称的。这道题可以使用递归和迭代（使用队列）来做，具体可参见代码

• java代码实现：

  import java.util.*;
  public class Solution {
  /**
   *
   * @param root TreeNode类
   * @return bool布尔型
   */
  public boolean isSymmetric (TreeNode root) {
      if(root == null)
          return true;
      return check(root.left, root.right);
  }
  
  //递归
  public boolean check(TreeNode left, TreeNode right){
      if(left == null && right == null)
          return true;
      if(left == null || right == null)
          return false;
      return left.val == right.val && check(left.left, right.right) && check(left.right, right.left);
  }
  
  //迭代
  public boolean isSymmetric (TreeNode root) {
      if(root == null)
          return true;
      Queue<TreeNode> queue = new LinkedList<>();
      queue.offer(root.left);
      queue.offer(root.right);
      while(!queue.isEmpty()){
          TreeNode left = queue.poll();
          TreeNode right = queue.poll();
          if(left == null && right == null)
              continue;
          if(left == null || right == null || left.val != right.val)
              return false;
          queue.offer(left.left);
          queue.offer(right.right);
          queue.offer(left.right);
          queue.offer(right.left);
      }
      return true;
  }
  }