# Checking if two binary trees are identical

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

Tree is defined as :

    /**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/


And my solution is

public boolean isSameTree(TreeNode p, TreeNode q) {

if(null == p && null == q){
return true;
}

if((null == p && null != q) || (null == q && null != p)){
return false;
}

if(p.val == q.val){
return isSameTree(p.left,q.left) ? isSameTree(p.right,q.right) :false;
} else {
return false;
}

}


Your code already does its job well. It can be shortened a bit:

public boolean isSameTree(TreeNode p, TreeNode q) {
if (p == null || q == null) {
return p == q;
}

return p.val == q.val
&& isSameTree(p.left, q.left)
&& isSameTree(p.right, q.right);
}


I flipped the operands of the == operators since having null in the right hand side reads more naturally.

I added spaces after the if and after the commas to follow the common formatting style.

• p == null && q == null can even be shorted to p == q in this case. Oct 30 '17 at 8:22
• @RolandIllig shorter code is not always better. I could not immediately determine why the p==q is correct here. So I would either comment it in, or be more verbose. btw I do really like the && in the return statement. Oct 30 '17 at 14:20