LeetCode 951. Flip Equivalent Binary Trees

方法一：递归

树的题，大部分都是递归。这道题的话，和same tree很像，无非就是加个判断看看是不是翻转的。

时间复杂度：T(n) = 4T(n/2), by Master Theorem, 时间复杂度 O(n^2)

空间复杂度：O(h)

class Solution {
public:bool flipEquiv(TreeNode* root1, TreeNode* root2) {if (root1==NULL && root2==NULL) return true;if (root1==NULL || root2==NULL) return false;// here root1 not null, root2 not nullif (root1->val!=root2->val) return false;return flipEquiv(root1->left, root2->left) && flipEquiv(root1->right, root2->right)|| flipEquiv(root1->left, root2->right) && flipEquiv(root1->right, root2->left);}
};

方法二：Serialize Tree

Serialize Tree方法很多，可以参考那一道题。这里就用preorder。

考虑到翻转，所以序列化的时候，dfs的时候，左右儿子中val小的先遍历，这样就算有翻转的情况，最后serialize的结果也是一样的。

class Solution {
public:bool flipEquiv(TreeNode* root1, TreeNode* root2) {return serialize(root1)==serialize(root2);}string serialize(TreeNode *root){if (root==NULL) return "#";string res=to_string(root->val);int left_val=root->left!=NULL?root->left->val:-1;int right_val=root->right!=NULL?root->right->val:-1;if (left_val<right_val)res += ' ' + serialize(root->left) + ' ' + serialize(root->right);elseres += ' ' + serialize(root->right) + ' ' + serialize(root->left);return res; }
};

时间复杂度：和遍历树类似，O(n)

空间复杂度：serialize后字符串需要占空间。

转载于:https://www.cnblogs.com/hankunyan/p/10117408.html

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