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.

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

思路：

该题主要使用递归的思路来做。递归比较两个子树。

• 递归到叶子结点，两个都为空返回true。
• 递归到叶子结点时，如果一个为空，另一个不为空则返回false。
• 两个结点都不为空但两个结点的值不同返回false。
• 递归比较两棵树的左子树、右子树。一棵树的左子树，另一棵树的右子树。一棵树的右子树，另一棵树的左子树。

代码：

public boolean flipEquiv(TreeNode root1, TreeNode root2) {
    if(root1 == null || root2 == null){
        return root1 == root2;
    }
    if(root1.val != root2.val){
        return false;
    }
    return (flipEquiv(root1.left, root2.left) && flipEquiv(root1.right, root2.right))
            || (flipEquiv(root1.left, root2.right) && flipEquiv(root1.right, root2.left));
}