For a binary tree T, we can define a flip operation as follows: choose any node, and swap the left and right child subtrees.
A binary tree X is flip equivalent to a binary tree Y if and only if we can make X equal to Y after some number of flip operations.
Given the roots of two binary trees root1
and root2
, return true
if the two trees are flip equivelent or false
otherwise.
Example 1:
Input: root1 = [1,2,3,4,5,6,null,null,null,7,8], root2 = [1,3,2,null,6,4,5,null,null,null,null,8,7] Output: true Explanation: We flipped at nodes with values 1, 3, and 5.
Example 2:
Input: root1 = [], root2 = [] Output: true
Example 3:
Input: root1 = [], root2 = [1] Output: false
Example 4:
Input: root1 = [0,null,1], root2 = [] Output: false
Example 5:
Input: root1 = [0,null,1], root2 = [0,1] Output: true
Constraints:
- The number of nodes in each tree is in the range
[0, 100]
. - Each tree will have unique node values in the range
[0, 99]
.
A:
递归比较
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */ class Solution { public: bool flipEquiv(TreeNode* root1, TreeNode* root2) { if(not root1 && not root2) return true; if(not root1 || not root2) return false; if(root1->val != root2->val) return false; int res1 = flipEquiv(root1->left, root2->left) && flipEquiv(root1->right, root2->right); return res1 || (flipEquiv(root1->left, root2->right) && flipEquiv(root1->right, root2->left)); } };
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