如果一个二叉树是另一棵树的子树

Question

I wrote this function to check if n2 is the subtree of n1 . I use recursion, but when I tested it using two trees, it showed me the wrong answer (expected true , but it actually returned false ).

I struggled for a while but still cannot say what's wrong.

private Boolean isSubTree(node n1, node n2){
    if(n1 == null)
        return false;
    if(n2 == null)
        return true;
    if(n1.data == n2.data){
        return isSubTree(n1.left,n2.left) && isSubTree(n2.right,n2.right);
    }
    else
        return isSubTree(n1.left, n2) || isSubTree(n1.right, n2);
}

Answer 1

Your edge cases are incorrect. In particular, when both n1 and n2 are null then that counts as a matching subtree, but if only one is null, then that counts as a mismatch.

Answer 2

I think it's easier to check n2's parent to see whether or not it's n1. if not, continue all the way until you find it; in which case you return true. the alternate is that all parents up to and including the root are not n1; in which case you return false. In other words, work your way up instead of checking all possible children and their children etc. of n1 to reach n2.

Answer 3

There are two cases to consider.

1. If the subtree has to be one of the nodes of the original tree then you can scan the tree for the matching subtree node.
2. If the subtree only has to have equal values, then you need to write a recursive tree equality function and apply it to all the leaves of the original tree in a second recursive function.

Answer 4

i agree with nabau ... its better to check weather n2's and n1's root node is same if we have data for parent in node

else we can check n2's root with node of every node in n1's

in java we can compare two objects are same or not
forEach all node in n1 equals root of n2 if any is node is equal they are same