本次選擇的題目是

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.

Solution：
運用遞迴的方法：

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

    private bool isSymmetricforLandR(TresNode l, TreeNode r){
        if(l == null || r == null){
            return l == r;
        }
        if(l.val != r.val){
            return
 
 false;
        }
        return isSymmetricforLandR(l.left, r.right) && isSymmetricforLandR(l.right, r.left);
    }
}

不用遞迴：

public boolean isSymmetric(TreeNode root) {
    if(root==null)  return true;

    Stack<TreeNode> stack = new Stack<TreeNode>();
    TreeNode left, right;
    if(root.left!=null){
        if(root.right==null) return false;
        stack.push(root.left);
        stack.push(root.right);
    }
    else if(root.right!=null){
        return false;
    }

    while(!stack.empty()){
        if(stack.size()%2!=0)   return false;
        right = stack.pop();
        left = stack.pop();
        if(right.val!=left.val) return false;

        if(left.left!=null){
            if(right.right==null)   return false;
            stack.push(left.left);
            stack.push(right.right);
        }
        else if(right.right!=null){
            return false;
        }

        if(left.right!=null){
            if(right.left==null)   return false;
            stack.push(left.right);
            stack.push(right.left);
        }
        else if(right.left!=null){
            return false;
        }
    }

    return true;
}