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.
Flipped Trees Diagram
Note:
- Each tree will have at most 100 nodes.
- Each value in each tree will be a unique integer in the range [0, 99].
Solution in C++
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int ChildSum(TreeNode *node) {
if (!node) return 0;
int sum=0;
if (node->left) sum += node->left->val;
if (node->right) sum += node->right->val;
return sum;
}
int ChildVal(TreeNode *node) {
if (node) return node->val;
return 0;
}
bool flipEquiv(TreeNode* root1, TreeNode* root2) {
if (!root1 && !root2) return true;
if (ChildVal(root1) != ChildVal(root2)) return false;
if (ChildSum(root1)!=ChildSum(root2)) return false;
if (root1->left) {
if (root2->left && ChildVal(root1->left)==ChildVal(root2->left))
return flipEquiv(root1->left, root2->left) && flipEquiv(root1->right, root2->right);
else
return flipEquiv(root1->left, root2->right) && flipEquiv(root1->right, root2->left);
}
else if (root1->right) {
if (root2->right && ChildVal(root1->right)==ChildVal(root2->right))
return flipEquiv(root1->right, root2->right) && flipEquiv(root1->left, root2->left);
else
return flipEquiv(root1->right, root2->left) && flipEquiv(root1->left, root2->right);
}
return true;
}
};