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ShusenTang 2020-02-05 17:00:42 +08:00
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solutions/101. Symmetric Tree.md
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@ -6,7 +6,9 @@
* 根的值相同;
* 且树一的左子树和树二的右子树互为镜像，树一的右子树和树二的左子树互为镜像。
## 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。
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。
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++
## 思路一
@ -14,13 +16,14 @@
class Solution {
private:
bool isSymmetricSameTree(TreeNode* p, TreeNode* q) { // 判断两棵树是否互为镜像
if(p == NULL && q==NULL) return true;
if(p == NULL || q == NULL || p -> val != q -> val) return false;
return isSymmetricSameTree(p -> left, q -> right) && isSymmetricSameTree(p -> right, q -> left);
if(!p && !q) return true;
if(!p || !q || p->val != q->val) return false;
return isSymmetricSameTree(p -> left, q -> right) && \
isSymmetricSameTree(p -> right, q -> left);
}
public:
bool isSymmetric(TreeNode* root) {
if(root == NULL) return true;
if(!root) return true;
return isSymmetricSameTree(root -> left, root -> right);
}
};
@ -30,23 +33,18 @@ public:
class Solution {
public:
bool isSymmetric(TreeNode *root) {
if (root == NULL) return true;
TreeNode *left, *right;
if (!root) return true;
TreeNode *left, *right;
queue<TreeNode*> q1, q2;
q1.push(root->left);
q2.push(root->right);
while (!q1.empty() && !q2.empty()){
left = q1.front();
q1.pop();
right = q2.front();
q2.pop();
if (NULL == left && NULL == right)
continue;
if (NULL == left || NULL == right)
return false;
if (left->val != right->val)
return false;
left = q1.front(); q1.pop();
right = q2.front(); q2.pop();
if (!left && !right) continue;
if (!left || !right || left->val != right->val) return false;
q1.push(left->left);
q1.push(left->right);
q2.push(right->right);