LeetCode 872. Leaf-Similar Trees

Description

https://leetcode.com/problems/leaf-similar-trees/

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].

Explanation

Traverse both trees and collect leaf node values. Compare if two trees leaf node values are the same.

Python Solution

# 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 leafSimilar(self, root1: TreeNode, root2: TreeNode) -> bool:
        
        sequence1 = []
        sequence2 = []
        
        self.traverse(root1, sequence1)
        
        self.traverse(root2, sequence2)
        
        return sequence1 == sequence2
        
        
    def traverse(self, root, results):
        if not root:
            return
        
        if not root.left and not root.right:
            results.append(root.val)
            return
        
        
        self.traverse(root.left, results)        
        self.traverse(root.right, results)
  • Time Complexity: O(N).
  • Space Complexity: O(N).

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