这两道题，大同小异。 我都是用BFS，在遍历的过程，判断结构是否相同/对称，值是否相同。

下面是AC代码：

1 /**  2      * Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).  3      * @param root  4      * @return  5      */  6     public boolean isSymmetricRecursively(TreeNode root){  7         if(root == null)  8             return true;  9         return isSymmetricR(root.left, root.right); 10     } 11     /** 12      * 采用迭代的方式，输入的是两个对称位置的节点，如果以这两个节点为root的子树是对称的，则 13      * 左边的左子树和右边的右子树相同，且左边的右子树和右边的左子树相同，且这两个子树的树根的值相等 14      * @param left 15      * @param right 16      * @return 17      */ 18     private boolean isSymmetricR(TreeNode left, TreeNode right){ 19         if(left== null && right == null) 20             return true; 21         if(left == null && right!=null || right == null && left!=null) 22             return false; 23         if(left.left == null && left.right == null && 24                 right.left == null && right.right == null) 25             return left.val == right.val; 26         boolean f1 = (left.val == right.val); 27         boolean l1 = isSymmetricR(left.left, right.right); 28         boolean r1 = isSymmetricR(left.right, right.left); 29         return f1&&l1&&r1; 30          31     } 32     /** 33      * Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center). 34      * solve it iteratively 35      * @param root 36      * @return 37      */ 38     public boolean isSymmetricIteratively(TreeNode root){ 39         if(root == null || (root.left == null && root.right == null)) 40             return true; 41         LinkedList
     
       qleft = new LinkedList
      
       (); 42         LinkedList
       
         qright = new LinkedList
        
         (); 43          44         if(root.left!=null && root.right == null || 45                 root.right!=null && root.left == null) 46             return false; 47         qleft.offer(root.left); 48         qright.offer(root.right); 49          50         while(!qleft.isEmpty() && !qright.isEmpty()){ 51             TreeNode left = qleft.poll(); 52             TreeNode right = qright.poll(); 53              54             if(left.val != right.val) 55                 return false; 56              57             if(left.right != null && right.left!=null){ 58                 qleft.offer(left.right); 59                 qright.offer(right.left); 60             }else if(left.right!= null && right.left == null || 61                     left.right == null && right.left !=null) 62                 return false; 63              64             if(right.right != null && left.left!=null){ 65                 qleft.offer(left.left); 66                 qright.offer(right.right); 67             }else if(right.right!= null && left.left == null || 68                     right.right == null && left.left !=null) 69                 return false; 70         } 71         return true; 72     } 73     /** 74      * by BFS, to check if they are equal or not 75      * Two binary trees are considered equal if they are structurally identical and the nodes have the same value. 76      * @param p 77      * @param q 78      * @return 79      */ 80     public boolean isSameTree(TreeNode p, TreeNode q){ 81         if(p == null && q == null) 82             return true; 83         if(p == null && q!=null || p!=null && q==null) 84             return false; 85         LinkedList
         
           pqueue = new LinkedList
          
           (); 86 LinkedList
           
             qqueue = new LinkedList
            
             (); 87 88 pqueue.offer(p); 89 qqueue.offer(q); 90 91 while(!pqueue.isEmpty() && !qqueue.isEmpty()){ 92 TreeNode pC = pqueue.poll(); 93 TreeNode qC = qqueue.poll(); 94 //have same value 95 if(pC.val!=qC.val) 96 return false; 97 //structurally identical or not 98 if(pC.left!=null && qC.left == null || 99 pC.left == null && qC.left!=null)100 return false;101 else if(pC.left!=null && qC.left!=null){102 pqueue.offer(pC.left);103 qqueue.offer(qC.left);104 }105 106 if(pC.right == null && qC.right != null || 107 pC.right!=null && qC.right == null)108 return false;109 else if(pC.right!=null && qC.right!=null){110 pqueue.offer(pC.right);111 qqueue.offer(qC.right);112 }113 }114 /**115 * there is no need116 * if(pqueue.isEmpty() && !qqueue.isEmpty() ||117 qqueue.isEmpty() && !pqueue.isEmpty())118 return false;119 */120 return true;121 }