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951. Flip Equivalent Binary Trees

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

DFS.

• Java
• C++
• Python
• Go

• /**
   * 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.
  # 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.
   * 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)
  }
  

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