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.
递归的解法：
新建一个递归调用自身的函数，输入是待比较的左结点和右结点，输出是对称判定结果：true or false；
1 class Solution { 2 public: 3 bool isSymmetric(TreeNode *root) { 4 if (!root) 5 return true; 6 7 return isSame(root->left, root->right); 8 9 } 10 11 bool isSame(TreeNode *node_l, TreeNode *node_r){ 12 if (!node_l && !node_r) 13 return true; 14 15 if (!node_l || !node_r) 16 return false; 17 18 if (node_l->val != node_r->val){ 19 return false; 20 } 21 22 return isSame(node_l->left, node_r->right) && 23 isSame(node_l->right, node_r->left); 24 } 25 26 };
迭代的解法：
新建两个栈用于存放待比较的左子树结点和右子树结点。每次迭代拿出两个栈中同位置元素进行比较，结束后删除此比较过的元素，并将其左右儿子压栈待比较。（还有只用一个栈的方法，并没有节省空间，略）
1 class Solution { 2 public: 3 bool isSymmetric(TreeNode *root) { 4 if (!root) 5 return true; 6 7 stack<TreeNode *> leftNodeStack; 8 stack<TreeNode *> rightNodeStack; 9 leftNodeStack.push(root->left); 10 rightNodeStack.push(root->right); 11 TreeNode *leftNode; 12 TreeNode *rightNode; 13 14 while (!leftNodeStack.empty()) { 15 leftNode = leftNodeStack.top(); 16 rightNode = rightNodeStack.top(); 17 leftNodeStack.pop(); 18 rightNodeStack.pop(); 19 20 if (!leftNode && !rightNode) { 21 continue; 22 } 23 24 if ((!leftNode && rightNode) || 25 (leftNode && !rightNode)) { 26 return false; 27 } 28 29 if (leftNode->val != rightNode->val) { 30 return false; 31 } 32 33 leftNodeStack.push(leftNode->left); 34 leftNodeStack.push(leftNode->right); 35 rightNodeStack.push(rightNode->right); // Notice 36 rightNodeStack.push(rightNode->left); 37 } 38 return true; 39 } 40 };