1 解題思想

這道題是上一道題的延伸版。
首先，題目的意思是說給了一顆樹，讓我們判斷他是否是對稱的(以根節點為中心，映象顛倒而成）

其實解題方式也很簡單，把根節點的左右節點開始，當做兩個獨立的子樹，判斷這兩個子樹的是否對稱

而判斷這兩個子樹是否對稱，就是在判斷兩個子樹的那個演算法那裡，把原來的左左對比，右右對比，改成左右對比和右左對比就可以了，稍稍改動一下位置就可以

2 原題

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

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

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

3 AC解

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
 

 /**
  * 當成兩棵樹遍歷檢查就可以了
  * */
public class Solution {
    public boolean check(TreeNode leftPart,TreeNode rightPart){
        if(leftPart==null && rightPart==null)
            return true;
        if(leftPart==null || rightPart==null)
            return false;
        if(leftPart.val != rightPart.val)
            return
 
 false;
        return check(leftPart.right,rightPart.left) && check(leftPart.left,rightPart.right);
    }
    public boolean isSymmetric(TreeNode root) {
        if(root == null)
            return true;
        return check(root.left,root.right);
    }
}