5. Same Tree

Description

Same Tree

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.

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

Constraints:

• The number of nodes in both trees is in the range [0, 100].
• -104 <= Node.val <= 104

Approach - Extension of Same Tree problem

class Solution {
    public boolean isSubtree(TreeNode root, TreeNode subRoot) {
        if (root == null)
            return false;
 
        if (isSameTree(root, subRoot))
            return true;
 
        return isSubtree(root.left, subRoot) || isSubtree(root.right, subRoot);        
    }
 
    public boolean isSameTree(TreeNode a, TreeNode b) {
        if (a == null && b == null)
            return true;
 
        if (a != null && b != null && a.val == b.val) {
            return isSameTree(a.right,b.right) && isSameTree(a.left,b.left);
        }
 
        return false;
    }
}

Approach

• Similar to depth but with 2 queues
• Check for scenario when either one tree is empty i.e. false and both tree are empty aka should return true
• Time: O(n) Space: O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        Queue<TreeNode> a = new LinkedList<>();
        Queue<TreeNode> b = new LinkedList<>();
        if (p != null) a.add(p);
        if (q != null) b.add(q);
        while(!a.isEmpty() && !b.isEmpty()) {
            for (int i = a.size() - 1; i >= 0; i--) {
                TreeNode x = a.poll(), y = b.poll();
                if (x == null && y == null) continue;
                if (x == null || y == null || x.val != y.val) 
	                return false;
                a.add(x.left);
                a.add(x.right);
                b.add(y.left);
                b.add(y.right);
            }
        }
        return (a.isEmpty() && b.isEmpty());
    }
}

Approach - Recursion

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        if (p == null && q == null) return true;
        if (p != null && q != null && p.val == q.val)
            return isSameTree(p.left,q.left) && isSameTree(p.right,q.right);
        return false;        
    }
}

---

Approach 1: Recursive DFS ($O(N)$ Time, $O(H)$ Space)

class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        // Base cases
        if (p == null && q == null) return true;
        if (p == null || q == null || p.val != q.val) return false;
 
        // Recursively check left and right subtrees
        return isSameTree(p.left, q.left) && isSameTree(p.right, q.right);
    }
}
 

Complexity

• Time Complexity: $O(N)$ — Visits each node in both trees at most once, where $N$ is the total number of nodes in the smaller tree.
• Space Complexity: $O(H)$ — Bounded by the height $H$ of the call stack ($O(\log N)$ for balanced trees, $O(N)$ for skewed trees).

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Approach 2: BFS Level-Order Traversal ($O(N)$ Time, $O(N)$ Space)

import java.util.ArrayDeque;
import java.util.Deque;
 
class Solution {
    public boolean isSameTree(TreeNode p, TreeNode q) {
        Deque<TreeNode> queue = new ArrayDeque<>();
        queue.offer(p);
        queue.offer(q);
 
        while (!queue.isEmpty()) {
            TreeNode n1 = queue.poll();
            TreeNode n2 = queue.poll();
 
            if (n1 == null && n2 == null) continue;
            if (n1 == null || n2 == null || n1.val != n2.val) return false;
 
            // Push corresponding children side-by-side
            queue.offer(n1.left);
            queue.offer(n2.left);
            queue.offer(n1.right);
            queue.offer(n2.right);
        }
 
        return true;
    }
}
 

Complexity

• Time Complexity: $O(N)$ — Each pair of nodes is pushed and polled from the queue once.
• Space Complexity: $O(N)$ — Queue holds at most one tree level’s nodes at any given time.

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Easy Memory Rule

“Both null? true. One null or values differ? false. Otherwise: recurse / queue left-with-left and right-with-right!”