[LC] 951. Flip Equivalent Binary Trees

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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].

Time: O(N)

/**
 * 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) {
        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);

    }
}

[LC] 951. Flip Equivalent Binary Trees

标签：wap   ber   return   public   style   operation   note   efi   write   

原文地址：https://www.cnblogs.com/xuanlu/p/13063083.html