Leetcode: Symmetric Tree

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



For example, this binary tree is symmetric:



    1

   / \

  2   2

 / \ / \

3  4 4  3

But the following is not:

    1

   / \

  2   2

   \   \

   3    3

Note:

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

难度：82. 这道题是树的题目，本质上还是树的遍历。这里无所谓哪种遍历方式，只需要对相应结点进行比较即可。一颗树对称其实就是看左右子树是否对称，一句话就是左同右，右同左，结点是对称的相等。不对称的条件有以下三个：（1）左边为空而右边不为空；（2）左边不为空而右边为空；（3）左边值不等于右边值。根据这几个条件在遍历时进行判断即可。

 1 /**

 2  * Definition for binary tree

 3  * public class TreeNode {

 4  *     int val;

 5  *     TreeNode left;

 6  *     TreeNode right;

 7  *     TreeNode(int x) { val = x; }

 8  * }

 9  */

10 public class Solution {

11     public boolean isSymmetric(TreeNode root) {

12         if (root == null) return true;

13         return check(root.left, root.right);

14     }

15     

16     public boolean check(TreeNode node1, TreeNode node2) {

17         if (node1 == null && node2 == null) return true;

18         else if (node1 == null && node2 != null) return false;

19         else if (node1 != null && node2 == null) return false;

20         else if (node1.val != node2.val) return false;

21         return check(node1.left, node2.right) && check(node1.right, node2.left);

22     }

23 }

Iterative的方式参考了网上的解法：

public boolean isSymmetric(TreeNode root) {

    if(root == null)

        return true;

    if(root.left == null && root.right == null)

        return true;

    if(root.left == null || root.right == null)

        return false;

    LinkedList<TreeNode> q1 = new LinkedList<TreeNode>();

    LinkedList<TreeNode> q2 = new LinkedList<TreeNode>();

    q1.add(root.left);

    q2.add(root.right);

    while(!q1.isEmpty() && !q2.isEmpty())

    {

        TreeNode n1 = q1.poll();

        TreeNode n2 = q2.poll();

        

        if(n1.val != n2.val)

            return false;

        if(n1.left == null && n2.right != null || n1.left != null && n2.right == null)

            return false;

        if(n1.right == null && n2.left != null || n1.right != null && n2.left == null)

            return false;

        if(n1.left != null && n2.right != null)

        {

            q1.add(n1.left);

            q2.add(n2.right);

        }

        if(n1.right != null && n2.left != null)

        {

            q1.add(n1.right);

            q2.add(n2.left);

        }            

    }

    return true;

}