Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
1
/
2 2
/ /
3 4 4 3
But the following is not:
1
/
2 2
3 3
Note: Bonus points if you could solve it both recursively and iteratively.
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
分析：这道题既可以用深度优先搜索DFS，也可以用广度优先搜索BFS。
思路：（1）深度优先搜索的思路为：
(1)传入root的左子树和右子树。如果两个都为NULL，则是对称的。
(2)如果两边都不为NULL并且两边的所对应的val相等，那就判断root->left->left和root->right->right是否对称，且判断root->left->right和root->right->left是否对称。。。
其余情况下return false;
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool func(TreeNode* left, TreeNode* right){ 13 if(left == NULL && right == NULL) 14 return true; 15 if(left != NULL && right != NULL && left->val == right->val) 16 return (func(left->right, right->left) && func(right->right, left->left)); 17 else 18 return false; 19 } 20 bool isSymmetric(TreeNode* root) { 21 if(root == NULL) 22 return true; 23 else 24 return func(root->left, root->right); 25 } 26 };
（2）用广度优先搜索来做:
建立两个队列，lq和rq，每次比较队首的两个元素是否相等。
lq队列的队首l出列后，将它的left和right分别入队；
rq队列的队首r出列后，将它的right和left分别入队。
因为判断是否对称是这样比较的：
l->left 与 r->right
l->right 与 r->left比较。
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */ 10 class Solution { 11 public: 12 bool isSymmetric(TreeNode* root) { 13 if(root == NULL) 14 return true; 15 queue<TreeNode*> lq, rq; 16 TreeNode *l, *r; 17 lq.push(root->left); 18 rq.push(root->right); 19 while(!lq.empty() && !rq.empty()){ 20 l = lq.front(); 21 r = rq.front(); 22 lq.pop(); 23 rq.pop(); 24 if(l == NULL && r == NULL) 25 continue; 26 if(l == NULL || r == NULL || l->val != r->val) 27 return false; 28 lq.push(l->left); 29 lq.push(l->right); 30 rq.push(r->right); 31 rq.push(r->left); 32 } 33 if(lq.empty() && rq.empty()) 34 return true; 35 else 36 return false; 37 38 } 39 };