[leetcode] 872. Leaf-Similar Trees

Description

Consider all the leaves of a binary tree, from left to right order, the values of those leaves form a leaf value sequence.

For example, in the given tree above, the leaf value sequence is (6, 7, 4, 9, 8).

Two binary trees are considered leaf-similar if their leaf value sequence is the same.

Return true if and only if the two given trees with head nodes root1 and root2 are leaf-similar.

Example 1:

Input: root1 = [3,5,1,6,2,9,8,null,null,7,4], root2 = [3,5,1,6,7,4,2,null,null,null,null,null,null,9,8] Output: true

Example 2:

Input: root1 = [1], root2 = [1] Output: true

Example 3:

Input: root1 = [1], root2 = [2] Output: false

Example 4:

Input: root1 = [1,2], root2 = [2,2] Output: true

Example 5:

Input: root1 = [1,2,3], root2 = [1,3,2] Output: false

Constraints:

The number of nodes in each tree will be in the range [1, 200].Both of the given trees will have values in the range [0, 200].

分析

题目的意思是：给定两个二叉树，然后判断两个二叉树的叶子结点是否相同。这道题最直观的思路是把两颗树的叶子结点都找出来，然后遍历判断一下就行了。

代码

# 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 solve(self,root,res): if(root is None): return if(root.left): self.solve(root.left,res) if(root.right): self.solve(root.right,res) if(root.left is None and root.right is None): res.append(root.val) def leafSimilar(self, root1: TreeNode, root2: TreeNode) -> bool: s1=[] s2=[] self.solve(root1,s1) self.solve(root2,s2) if(len(s1)!=len(s2)): return False for i in range(len(s1)): if(s1[i]!=s2[i]): return False return True