leetcode951. Flip Equivalent Binary Trees

题目链接
题目：给出两棵树，问能不能通过翻转（互换）左右孩子节点而互相转化
思路：递归进行比较，第一棵树的左孩子和第二棵数的左孩子，第一棵树的右孩子和第二棵数的右孩子或者第一棵树的左孩子和第二棵数的右孩子，第一棵树的右孩子和第二棵数的左孩子。

/*** Definition for a binary tree node.* public class TreeNode {*     int val;*     TreeNode left;*     TreeNode right;*     TreeNode(int x) { val = x; }* }*/
class Solution {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);}
}

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