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update code of solution2

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solutions/101. Symmetric Tree.md
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@ -8,6 +8,8 @@
## 思路二: 非递归 ## 思路二: 非递归
采用类似层次遍历的方法。对于左子树从上往下**从左往右**层次遍历，对于右子树从上往下**从右往左**层次遍历，遍历过程中进行比较。注意遍历过程中如果某个节点的孩子为空也应将其孩子入队。 采用类似层次遍历的方法。对于左子树从上往下**从左往右**层次遍历，对于右子树从上往下**从右往左**层次遍历，遍历过程中进行比较。注意遍历过程中如果某个节点的孩子为空也应将其孩子入队。
层次遍历用的队列，其实用栈也是可以的，代码几乎一样，这样就是分别用**根右左**和**根左右**来遍历左右子树，同理空孩子也需要入栈。
两种思路都相当于遍历一遍树，所以两种思路的时空复杂度均为O(n)。 两种思路都相当于遍历一遍树，所以两种思路的时空复杂度均为O(n)。
# C++ # C++
@ -35,20 +37,28 @@ public:
bool isSymmetric(TreeNode* root) { bool isSymmetric(TreeNode* root) {
if(!root) return true; if(!root) return true;
TreeNode *left, *right; queue<TreeNode *>container1, container2; // use queue
queue<TreeNode*> q1, q2; // stack<TreeNode *>container1, container2; // use stack
q1.push(root->left);
q2.push(root->right);
while (!q1.empty() && !q2.empty()){ container1.push(root -> left);
left = q1.front(); q1.pop(); container2.push(root -> right);
right = q2.front(); q2.pop();
if (!left && !right) continue; TreeNode *p1, *p2;
if (!left || !right || left->val != right->val) return false; while(!container1.empty() && !container2.empty()){
q1.push(left->left); p1 = container1.front(); p2 = container2.front(); // use queue
q1.push(left->right); // p1 = container1.top(); p2 = container2.top(); // use stack
q2.push(right->right);
q2.push(right->left); container1.pop(); container2.pop();
if(p1 && p2){
if(p1 -> val != p2 -> val) return false;
container1.push(p1 -> left);
container1.push(p1 -> right);
container2.push(p2 -> right);
container2.push(p2 -> left);
}
else if(!p1 && !p2) continue;
else return false;
} }
return true; return true;
} }