leetcode 101. 对称二叉树（递归与非递归）

	/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
/*class Solution {
public:
    bool dfs(TreeNode*r1,TreeNode*r2)//递归
    {
      if(!r1||!r2)return !r1&&!r2;
      return r1->val==r2->val&&dfs(r1->left,r2->right)&&dfs(r1->right,r2->left)
     
    }
    bool isSymmetric(TreeNode* root) {    
      if(!root)return true;
      return dfs(root->left,root->right);   
    }
};*/

class Solution {
public:
  
    bool isSymmetric(TreeNode* root) {    //非递归模拟递归
      if(!root)return true;
      stack<TreeNode*>left,right;
      auto l=root->left,r=root->right;
      while(l||r||left.size()||right.size())
      {
        while(l&&r)
        {
          left.push(l),right.push(r);//左孩子右孩子分别压入两个栈
          l=l->left,r=r->right;
        }
        if(l||r)return false;//如果跳出循环还有不为空说明非镜像
        
        l=left.top(),r=right.top();
        left.pop(),right.pop();
        if(l->val!=r->val)return false;//判断栈顶元素
        l=l->right,r=r->left;//加入右左孩子；
      }
      
      return true;
      
    }
};