题目：101. Symmetric Tree

判断一棵树是否是镜像的。简单的DFS题。

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.

核心就是：对称的两边node都不为空（否则就false）的情况下，分别递归判断左边的左node／右边的右node && 左边的右node／右边的左node

class Solution {
    public boolean isSymmetric(TreeNode root) {
        if(root == null){return true; }
        return isMirror(root.left, root.right);
      
    }
    
    public boolean isMirror(TreeNode left, TreeNode right){
        if(left == null && right == null) {
            return true;}
        else if((left == null && right != null) || 
            (left != null && right == null) ||
            (left.val != right.val)  ){
            return false;
        }else{
            return isMirror(left.left, right.right) && isMirror(left.right, right.left) ;
            
        }        
    }
}

可以写的更简洁：

class Solution {
    public boolean isSymmetric(TreeNode root) {
        return dfs(root, root);
    }
    
    private boolean dfs(TreeNode left, TreeNode right) {
        if (left == null || right == null) return left == right;
        
        if (left.val != right.val) return false;
        
        return dfs(left.right, right.left) && dfs(left.left, right.right);
    }
}