Question

951. 翻转等价二叉树

English Version

题目描述

我们可以为二叉树 T 定义一个 翻转操作 ，如下所示：选择任意节点，然后交换它的左子树和右子树。

只要经过一定次数的翻转操作后，能使 X 等于 Y，我们就称二叉树 X _翻转 等价 _于二叉树 Y。

这些树由根节点 root1 和 root2 给出。如果两个二叉树是否是_翻转 等价 _的函数，则返回 true ，否则返回 false 。

 

示例 1：

输入：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]
输出：true
解释：我们翻转值为 1，3 以及 5 的三个节点。

示例 2:

输入: root1 = [], root2 = []
输出: true

示例 3:

输入: root1 = [], root2 = [1]
输出: false

 

提示：

• 每棵树节点数在 [0, 100] 范围内
• 每棵树中的每个值都是唯一的、在 [0, 99] 范围内的整数

解法

方法一

# 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)
}