Question

951. Flip Equivalent Binary Trees

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Description

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].

Solutions

Solution 1

# Definition for a binary tree node.
# class TreeNode:
#   def __init__(self, val=0, left=None, right=None):
#     self.val = val
#     self.left = left
#     self.right = right
class Solution:
  def flipEquiv(self, root1: Optional[TreeNode], root2: Optional[TreeNode]) -> bool:
    def dfs(root1, root2):
      if root1 == root2 or (root1 is None and root2 is None):
        return True
      if root1 is None or root2 is None or root1.val != root2.val:
        return False
      return (dfs(root1.left, root2.left) and dfs(root1.right, root2.right)) or (
        dfs(root1.left, root2.right) and dfs(root1.right, root2.left)
      )

    return dfs(root1, root2)

/**
 * 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));
  }
}

/**
 * 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) {
    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));
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func flipEquiv(root1 *TreeNode, root2 *TreeNode) bool {
  var dfs func(root1, root2 *TreeNode) bool
  dfs = func(root1, root2 *TreeNode) bool {
    if root1 == root2 || (root1 == nil && root2 == nil) {
      return true
    }
    if root1 == nil || root2 == nil || 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))
  }
  return dfs(root1, root2)
}