LeetCode题解(0872)：判断二叉树的叶是否相似(Python)

题目：原题链接（简单）

解法时间复杂度空间复杂度执行用时Ans 1 (Python)$O(T_1+T_2)$ : $T_1$和$T_2$分别为两树的深度$O(T_1+T_2)$ : $T_1$和$T_2$分别为两树的深度40ms (84.19%)Ans 2 (Python)$O(T_1+T_2)$ : $T_1$和$T_2$分别为两树的深度$O(T_1+T_2)$ : $T_1$和$T_2$分别为两树的深度44ms (65.15%)Ans 3 (Python)

LeetCode的Python执行用时随缘，只要时间复杂度没有明显差异，执行用时一般都在同一个量级，仅作参考意义。

解法一：

def leafSimilar(self, root1: TreeNode, root2: TreeNode) -> bool: def helper(node): if not node: return [] if node.left and node.right: return helper(node.left) + helper(node.right) elif node.left: return helper(node.left) elif node.right: return helper(node.right) else: return [node.val] return helper(root1) == helper(root2)

解法二：

def leafSimilar(self, root1: TreeNode, root2: TreeNode) -> bool: def helper(node): if node: if not node.left and not node.right: yield node.val yield from helper(node.left) yield from helper(node.right) return list(helper(root1)) == list(helper(root2))