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<TreeNode> qleft = new LinkedList<TreeNode>(); 42 LinkedList<TreeNode> qright = new LinkedList<TreeNode>(); 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<TreeNode> pqueue = new LinkedList<TreeNode>(); 86 LinkedList<TreeNode> qqueue = new LinkedList<TreeNode>(); 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 need 116 * if(pqueue.isEmpty() && !qqueue.isEmpty() || 117 qqueue.isEmpty() && !pqueue.isEmpty()) 118 return false; 119 */ 120 return true; 121 }