951. Flip Equivalent Binary Trees

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 equivalent 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

 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].

C++

class Solution {
public:
    bool flipEquiv(TreeNode* root1, TreeNode* root2) {
        return dfs(root1, root2);
    }

    bool dfs(TreeNode* root1, TreeNode* root2) {
        if (root1 == root2 || (!root1 && !root2)) return true;
        if (!root1 || !root2 || root1->val != root2->val) return false;
        return (dfs(root1->left, root2->left) && dfs(root1->right, root2->right)) || (dfs(root1->left, root2->right) && dfs(root1->right, root2->left));
    }
};

JAVA

• /**
   * Definition for a binary tree node.
   * public class TreeNode {
   *     int val;
   *     TreeNode left;
   *     TreeNode right;
   *     TreeNode() {}
   *     TreeNode(int val) { this.val = val; }
   *     TreeNode(int val, TreeNode left, TreeNode right) {
   *         this.val = val;
   *         this.left = left;
   *         this.right = right;
   *     }
   * }
   */
  class Solution {
      public boolean flipEquiv(TreeNode root1, TreeNode root2) {
          return dfs(root1, root2);
      }
  
      private boolean dfs(TreeNode root1, TreeNode root2) {
          if (root1 == root2 || (root1 == null && root2 == null)) {
              return true;
          }
          if (root1 == null || root2 == null || root1.val != root2.val) {
              return false;
          }
          return (dfs(root1.left, root2.left) && dfs(root1.right, root2.right))
              || (dfs(root1.left, root2.right) && dfs(root1.right, root2.left));
      }
  }

PYTHON

class Solution:
    def flipEquiv(self, root1, root2):
        return self.dfs(root1, root2)

    def dfs(self, root1, root2):
        if root1 == root2 or (not root1 and not root2):
            return True
        if not root1 or not root2 or root1.val != root2.val:
            return False
        return (self.dfs(root1.left, root2.left) and self.dfs(root1.right, root2.right)) or \
               (self.dfs(root1.left, root2.right) and self.dfs(root1.right, root2.left))