951. Flip Equivalent Binary Trees**
https://leetcode.com/problems/flip-equivalent-binary-trees/
題目描述
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.
Write a function that determines whether two binary trees are flip equivalent. The trees are given by root nodes root1 and root2.
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.
Note:
- Each tree will have at most 100 nodes.
- Each value in each tree will be a unique integer in the range
[0, 99].
C++ 實現 1
其中
if (!root1 || !root2) return root1 == root2;
不必多說. 在兩個 root 均不爲空的情況下, 如果它們的值不相等, 那麼就返回 false.
而如果它們的值相等, 就需要判斷左右子樹是不是相等, 或者翻轉後左右子樹是否相等.
public:
bool flipEquiv(TreeNode* root1, TreeNode* root2) {
if (!root1 || !root2) return root1 == root2;
if (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));
}
};