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🟩 Same Tree (#100)

🔗 Problem Link​

📋 Problem Statement​

Given the roots of two binary trees p and q, 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.

💡 Examples​

Example 1​

Input: p = [1,2,3], q = [1,2,3]
Output: true

Example 2​

Input: p = [1,2], q = [1,null,2]
Output: false

Example 3​

Input: p = [1,2,1], q = [1,1,2]
Output: false

🔑 Key Insights & Approach​

Core Observation: Recursively check if current nodes match and subtrees match.

Why DFS Recursion?

  • Natural recursive structure
  • O(n) time, O(h) space
  • Clean base cases

Approaches:

  1. Recursive DFS: O(n) time, O(h) space - optimal
  2. Iterative BFS: O(n) time, O(n) space

Pattern: "Tree Comparison" recursive pattern.

🐍 Solution: Python​

Approach: Recursive DFS​

Time Complexity: O(n) | Space Complexity: O(h)

from typing import Optional

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: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
        # Both null
        if not p and not q:
            return True

        # One null, one not
        if not p or not q:
            return False

        # Values don't match
        if p.val != q.val:
            return False

        # Check both subtrees
        return self.isSameTree(p.left, q.left) and self.isSameTree(p.right, q.right)

Approach 2: Iterative BFS​

Time Complexity: O(n) | Space Complexity: O(n)

from collections import deque
from typing import Optional

class Solution:
    def isSameTree(self, p: Optional[TreeNode], q: Optional[TreeNode]) -> bool:
        queue = deque([(p, q)])

        while queue:
            node1, node2 = queue.popleft()

            if not node1 and not node2:
                continue

            if not node1 or not node2:
                return False

            if node1.val != node2.val:
                return False

            queue.append((node1.left, node2.left))
            queue.append((node1.right, node2.right))

        return True

🔵 Solution: Golang​

Approach: Recursive DFS​

Time Complexity: O(n) | Space Complexity: O(h)

type TreeNode struct {
    Val   int
    Left  *TreeNode
    Right *TreeNode
}

func isSameTree(p *TreeNode, q *TreeNode) bool {
    // Both null
    if p == nil && q == nil {
        return true
    }

    // One null
    if p == nil || q == nil {
        return false
    }

    // Values don't match
    if p.Val != q.Val {
        return false
    }

    // Check subtrees
    return isSameTree(p.Left, q.Left) && isSameTree(p.Right, q.Right)
}

📚 References​