golang 解法, 执行用时: 0 ms, 内存消耗: 2.4 MB, 提交时间: 2022-11-27 11:56:44

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func flipEquiv(root1 *TreeNode, root2 *TreeNode) bool {
    if root1 == root2 {
        return true
    }
    if root1 == nil || root2 == nil || root1.Val != root2.Val {
        return false
    }
    
    return (flipEquiv(root1.Left, root2.Left) && flipEquiv(root1.Right, root2.Right)) || 
            (flipEquiv(root1.Right, root2.Left) && flipEquiv(root1.Left, root2.Right))
    
}

python3 解法, 执行用时: 40 ms, 内存消耗: 15 MB, 提交时间: 2022-11-27 11:54:05

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flipEquiv(self, root1, root2):
        def dfs(node):
            if node:
                yield node.val
                L = node.left.val if node.left else -1
                R = node.right.val if node.right else -1
                if L < R:
                    yield from dfs(node.left)
                    yield from dfs(node.right)
                else:
                    yield from dfs(node.right)
                    yield from dfs(node.left)
                yield '#'

        return all(x == y for x, y in itertools.zip_longest(
            dfs(root1), dfs(root2)))

java 解法, 执行用时: 0 ms, 内存消耗: 39.4 MB, 提交时间: 2022-11-27 11:53:46

/**
 * 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));
    }
}

python3 解法, 执行用时: 32 ms, 内存消耗: 15.1 MB, 提交时间: 2022-11-27 11:53:29

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution(object):
    def flipEquiv(self, root1, root2):
        if root1 is root2:
            return True
        if not root1 or not root2 or root1.val != root2.val:
            return False

        return (self.flipEquiv(root1.left, root2.left) and
                self.flipEquiv(root1.right, root2.right) or
                self.flipEquiv(root1.left, root2.right) and
                self.flipEquiv(root1.right, root2.left))