algorithm - 将后序二叉树遍历索引转换为层序(广度优先)索引

Question

假设一个完整的二叉树，每个节点都可以用它在给定的树遍历算法中出现的位置来寻址。

例如，一棵高度为 3 的简单完整树的节点索引如下所示:

广度优先(又名水平顺序):

     0
   /   \
  1     2
 /  \  /  \
3   4  5  6

后订单部优先:

     6
   /   \
  2     5
 /  \  /  \
0   1  3  4

给出树的高度和后序遍历中的索引。

我如何根据这些信息计算广度优先索引？

Answer 1

我认为它必须迭代/递归计算。话虽如此，有人会在 37 秒内通过简单的单行计算来对我投反对票。尽管如此，它可以通过递归思考来解决。考虑深度优先后序遍历的简单树(基于 1):

   3
  / \
 1   2

从递归的角度来看，这就是您需要考虑的全部内容。您位于子树 (3) 的根部、子树 (1) 的左侧部分或右侧 (2)。如果你在根部，那么你就完成了。否则，左右子树相同，右子树中的后序遍历索引等于对应的左子树索引+子树中的节点数。

以下代码在 O(log n) 中执行此计算。对于深度为 10(1023 个节点)的树，它最多计算 10 次迭代(递归)的索引值。

它跟踪给定节点的深度，并根据它是处理左子树还是右子树来计算广度优先行位置。请注意，这使用基于 1 的索引值。我发现用这些术语来思考它更简单(深度为 2 的树中有 3 个节点，后序遍历中最顶层的节点是 3)。另请注意，它认为树深度为 1 时只有一个节点(我不确定这是否是典型约定)。

// Recursively compute the given post-order traversal index's position
// in the tree:  Its position in the given level and its depth in the tree.
void ComputePos( int treedepth, int poindex, int *levelposition, int *nodedepth )
{
    int nodes;
    int half;

    // compute number of nodes for this depth.
    assert( treedepth > 0 );
    nodes = ( 1 << ( treedepth )) - 1;
    half = nodes / 2;   // e.g., 7 / 2 = 3

    //printf( "poindex = %3d, Depth = %3d, node count = %3d", poindex, treedepth, nodes );

    (*nodedepth)++;

    if ( poindex == nodes ) {
        // This post-order index value is the root of this subtree
        //printf( "  Root\n" );
        return;
    }
    else if ( poindex > half ) {
        // This index is in the right subtree
        //printf( "  Right\n" );
        poindex -= half;
        *levelposition = 2 * *levelposition + 1;
    }
    else {
        // Otherwise it must be in the left subtree
        //printf( "  Left\n" );
        *levelposition = 2 * *levelposition;
    }

    treedepth -= 1;
    ComputePos( treedepth, poindex, levelposition, nodedepth );
}

int main( int argc, char* argv[] )
{
    int levelposition = 0;   // the 0-based index from the left in a given level
    int nodedepth = 0;  // the depth of the node in the tree
    int bfindex;
    int treedepth = atoi( argv[1] );   // full depth of the tree (depth=1 means 1 node)
    int poindex = atoi( argv[2] ); // 1-based post-order traversal index

    ComputePos( treedepth, poindex, &levelposition, &nodedepth );

    //printf( "ComputePos( %d, %d ) = %d, %d\n", treedepth, poindex, levelposition, nodedepth );

    // Compute the breadth-first index as its position in its current
    // level plus the count of nodex in all the levels above it.
    bfindex = levelposition + ( 1 << ( nodedepth - 1 ));
    printf( "Post-Order index %3d = breadth-first index %3d\n", poindex, bfindex );

    return 0;
}

这是为以下树(深度 4)计算的值，它显示了后序遍历索引值(从 1 开始)。

            15
          /    \
         /      \
        /        \
       /          \
      /            \
     7             14     
    / \            / \    
   /   \          /   \   
  3     6       10    13  
 /\    / \      /\    / \ 
1  2  4   5    8  9  11  12


[C:\tmp]for /l %i in (1,1,15) do po2bf 4 %i
Post-Order index   1 = breadth-first index   8
Post-Order index   2 = breadth-first index   9
Post-Order index   3 = breadth-first index   4
Post-Order index   4 = breadth-first index  10
Post-Order index   5 = breadth-first index  11
Post-Order index   6 = breadth-first index   5
Post-Order index   7 = breadth-first index   2
Post-Order index   8 = breadth-first index  12
Post-Order index   9 = breadth-first index  13
Post-Order index  10 = breadth-first index   6
Post-Order index  11 = breadth-first index  14
Post-Order index  12 = breadth-first index  15
Post-Order index  13 = breadth-first index   7
Post-Order index  14 = breadth-first index   3
Post-Order index  15 = breadth-first index   1