Leetcode 101 Symmetric Tree

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

这是一个镜像树的问题 注意姿势左右对称 也就是最左等于最右 

我写代码怎么就罗里吧嗦

/** * 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) return true;        return sameChild(root.left,root.right);           }        public boolean sameChild(TreeNode left,TreeNode right){               if((left==null && right!=null)||(left!=null && right==null))           return false;          if(left==null && right==null)//这些乱七八糟的if啊。。可以统一写啊啊啊= =           return true;                       if(left.val == right.val){            return sameChild(left.left, right.right) && sameChild(left.right, right.left);          }        else{            return false;        } 

            //if (!lt && !rt) return true;  举个栗子而已= =        //if (lt && !rt || !lt && rt || lt->val != rt->val) return false;          //return isSymmetric(lt->left, rt->right) &&isSymmetric(lt->right, rt->left); 

    }        }

贴一个先序遍历的做法= =

    bool isSymmetric(TreeNode *root)       {          if (!root || !root->left && !root->right) return true;          TreeNode *t1 = root->left, *t2 = root->right;          if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;            stack<TreeNode *> s1, s2;  //弄俩栈        s1.push(t1), s2.push(t2); //分别把root的左右孩子压进去         bool flag = false;          while (!s1.empty() && !s2.empty())  //嗯哼 栈空结束        {              if (!flag && (t1->left || t2->right))  //左孩子的左孩子 右孩子的右孩子            {                  s1.push(t1), s2.push(t2);                  t1 = t1->left, t2 = t2->right;                  if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;                  s1.push(t1), s2.push(t2);              }              else if (t1->right || t2->left)  //如果到了这一步说明左孩子没有左孩子 右孩子没有右孩子            {                  t1 = t1->right, t2 = t2->left;                  if (t1&&!t2 || !t1&&t2 || t1->val!=t2->val) return false;                  flag = false;              }              else  //到这了说明t1 t2都是叶子节点了所以要开始退栈            {                  t1 = s1.top(), t2 = s2.top();                  s1.pop(), s2.pop();                  flag = true;              }          }          return s1.empty() && s2.empty();      }