Interview Question


Country: United States




Comment hidden because of low score. Click to expand.
1
of 1 vote

Node * GetCommon(Node * node, Node * a, Node * b)
{
    if(node == 0){ return 0; }
    if(node == a){ return a; }
    if(node == b){ return b; }

    Node * l = GetCommon(node->left, a, b);
    Node * r = GetCommon(node->right,a,b);
    if(l && r){ return node; }
    return l ? l : r;
}

- Anonymous October 22, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

Your method has O(N) complexity in the worst situation, while there are methods with a logN complexity.

- uuuouou November 30, 2013 | Flag
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0
of 0 vote

Since you didn't mention anything, I will assume its a general tree. In that case, one of the ways is the following.

1. Print out the path from root to each node.
2. Find the common substring to this path.
3. The last node is the least common ancestor.

There are various implementations to this method including Dynamic Programming. (Due to no link restriction, I can't give a relevant link but a simple google search should do it)

- Hingle McCringleBerry October 21, 2013 | Flag Reply
Comment hidden because of low score. Click to expand.
0
of 0 votes

are you mad are what ..given tree as binary tree...,

- zxcvb March 13, 2014 | Flag


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