4
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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;
    }

}

Please suggest any improvements.

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4
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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.

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  • 1
    \$\begingroup\$ p == null && q == null can even be shorted to p == q in this case. \$\endgroup\$ – RoToRa Oct 30 '17 at 8:22
  • 1
    \$\begingroup\$ @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. \$\endgroup\$ – RobAu Oct 30 '17 at 14:20

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