LeetCode 951. Flip Equivalent Binary Trees

红太狼 2021-12-16 23:41 113阅读 0赞

原题链接在这里：[https://leetcode.com/problems/flip-equivalent-binary-trees/][https_leetcode.com_problems_flip-equivalent-binary-trees]

题目：

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.

Note:

1.  Each tree will have at most `100` nodes.
2.  Each value in each tree will be a unique integer in the range `[0, 99]`.

题解：

Compare root1 and root2, see if it is equal first.

If yes, then

Case 1: root1.left equivalent to root2.left && root1.right equivalent to root2.right

Case 2: root1.left equivalent to root2.right && root1.right equivalent to root2.left.

Either case is true, then return true.

Time Complexity: O(n^2). T(n) = 4\*T(n/2) + 1. Master Theorem, T(n) = O(n^2).

Space: O(h).

AC Java:

1 /**
     2  * Definition for a binary tree node.
     3  * public class TreeNode {
     4  *     int val;
     5  *     TreeNode left;
     6  *     TreeNode right;
     7  *     TreeNode(int x) { val = x; }
     8  * }
     9  */
    10 class Solution {
    11     public boolean flipEquiv(TreeNode root1, TreeNode root2) {
    12         if(root1 == null && root2 == null){
    13             return true;
    14         }
    15         
    16         if(root1 == null || root2 == null){
    17             return false;
    18         }
    19         
    20         if(root1.val != root2.val){
    21             return false;
    22         }
    23         
    24         return (flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right))
    25             || (flipEquiv(root1.left, root2.right) && flipEquiv(root1.right, root2.left));
    26     }
    27 }

转载于:https://www.cnblogs.com/Dylan-Java-NYC/p/11072771.html

[https_leetcode.com_problems_flip-equivalent-binary-trees]: https://leetcode.com/problems/flip-equivalent-binary-trees/