The full-text may be used and/or reproduced, and given to third parties in any format or medium, without prior permission or charge, for personal research or study, educational, or not-for-pro t purposes provided that:• a full bibliographic reference is made to the original source • a link is made to the metadata record in DRO • the full-text is not changed in any way The full-text must not be sold in any format or medium without the formal permission of the copyright holders.Please consult the full DRO policy for further details. Abstract. Let G be a simple undirected graph on n vertices with maximum degree ∆. Brooks' Theorem states that G has a ∆-colouring unless G is a complete graph, or a cycle with an odd number of vertices. To recolour G is to obtain a new proper colouring by changing the colour of one vertex. We show that from a k-colouring, k > ∆, a ∆-colouring of G can be obtained by a sequence of O(n 2 ) recolourings using only the original k colours unless -G is a complete graph or a cycle with an odd number of vertices, or -k = ∆ + 1, G is ∆-regular and, for each vertex v in G, no two neighbours of v are coloured alike. We use this result to study the reconfiguration graph R k (G) of the kcolourings of G. The vertex set of R k (G) is the set of all possible kcolourings of G and two colourings are adjacent if they differ on exactly one vertex. It is known that -if k ≤ ∆(G), then R k (G) might not be connected and it is possible that its connected components have superpolynomial diameter, -if k ≥ ∆(G) + 2, then R k (G) is connected and has diameter O(n 2 ). We complete this structural classification by settling the missing case:-if k = ∆(G) + 1, then R k (G) consists of isolated vertices and at most one further component which has diameter O(n 2 ). We also describe completely the computational complexity classification of the problem of deciding whether two k-colourings of a graph G of maximum degree ∆ belong to the same component of R k (G) by settling the case k = ∆(G) + 1. The problem is -O(n 2 ) time solvable for k = 3, -PSPACE-complete for 4 ≤ k ≤ ∆(G), -O(n) time solvable for k = ∆(G) + 1, -O(1) time solvable for k ≥ ∆(G) + 2 (the answer is always yes).