101. Symmetric Tree | 判断二叉树是否为镜像二叉树

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本文介绍了一种检查二叉树是否为中心对称的方法，通过递归方式验证每个节点的左右子树是否为镜像对称。文章提供了一个具体的实现示例。

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Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).

For example, this binary tree [1,2,2,3,4,4,3] is symmetric:

But the following [1,2,2,null,3,null,3] is not:

Note:
Bonus points if you could solve it both recursively and iteratively.

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/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
   	public boolean isSymmetric(TreeNode root) {
		if (root == null || root.left == null && root.right == null) {
			return true;
		}

		else if (root.left != null && root.right != null) {
			return isSymmetricHelper(root.left, root.right);
		} else {
			return false;
		}

	}

	public boolean isSymmetricHelper(TreeNode left, TreeNode right) {
		if (left == null && right == null) {
			return true;
		} else if (left == null || right == null) {
			return false;
		}
		return left.val == right.val && isSymmetricHelper(left.left, right.right)
				&& isSymmetricHelper(left.right, right.left);
	}
}