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# 951.Flip-Equivalent-Binary-Trees

## 题目描述

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

## 代码

### Approach #1 Recursion

Time: O(mini(N1, N2)) && Space: O(min(H1, H2))
/**
* 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 == root2) return true;
if (root1 == null || root2 == null || 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));
}
}

### Approach 2: Canonical Traversal

Time: O(N1 + N2)
class Solution {
public boolean flipEquiv(TreeNode root1, TreeNode root2) {
List<Integer> vals1 = new ArrayList();
List<Integer> vals2 = new ArrayList();
dfs(root1, vals1);
dfs(root2, vals2);
return vals1.equals(vals2);
}
public void dfs(TreeNode node, List<Integer> vals) {
if (node != null) {
int L = node.left != null ? node.left.val : -1;
int R = node.right != null ? node.right.val : -1;
if (L < R) {
dfs(node.left, vals);
dfs(node.right, vals);
} else {
dfs(node.right, vals);
dfs(node.left. vals);
}