5. 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;        
    }
}