100. Same Tree

Problem

Given two binary trees, write a function to check if they are the same or not.

Two binary trees are considered the same if they are structurally identical and the nodes have the same value.

Example 1:

Input:     1         1
          / \       / \
         2   3     2   3

        [1,2,3],   [1,2,3]

Output: true

Example 2:

Input:     1         1
          /           \
         2             2

        [1,2],     [1,null,2]

Output: false

Example 3:

Input:     1         1
          / \       / \
         2   1     1   2

        [1,2,1],   [1,1,2]

Output: false

Discussion

A DFS algorithm would be able to scan down both tree at the same time, and pointing out difference when spotted.

Solution

A simple DFS algorithm on both tree would works. Note that after we compare the value for both nodes, passing child node to recursion is sufficient enough for scanning.

# 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 isSameTree(self, p: TreeNode, q: TreeNode) -> bool:

        if not p and not q:
            return True

        if not p or not q:
            return False

        if p.val != q.val:
            return False

        return self.isSameTree(p.right, q.right) and \
            self.isSameTree(p.left, q.left)

Complexity Analysis

• Time Complexity: O(n), for simple DFS scanning the whole tree
• Space Complexity: O(n), for simple DFS holding all nodes in recursion. Note that for perfect balanced binary tree, space complexity would be just O(log(n)) as the depth of the tree would be log(n).