Microsoft interview question
BFS on a binary tree
Interview Answer
Anonymous
Aug 6, 2010
fun bfs(t,k) if (t is null) return not-found if (t.key is k) return t if (t is leaf) else r = bfs(t.left,k) if (r is not-found) r = bfs(t.right,k) return r
BFS on a binary tree
Anonymous
fun bfs(t,k) if (t is null) return not-found if (t.key is k) return t if (t is leaf) else r = bfs(t.left,k) if (r is not-found) r = bfs(t.right,k) return r
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