父级数组n-ary树的逐级遍历？

Question

Given an n-ary tree stored in a parent array, with the children stored in an array of pointers to arrays where the first value is the number of children:

(childArray[2][0] shows that node 2 has 2 children, childArray[2][1] shows that its first child is 5, etc.)

parentArray = {3, 0, 3, -1, 3, 2, 2};
childArray = {{1, 1}, {0}, {2, 5, 6}, {3, 0, 2, 4}, {0}, {0}, {0}};

produces a tree that looks like this:

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

Using a queue, how can I output the tree level by level like so:

Level 1: 3

Level 2: 0, 2, 4

Level 3: 1, 5, 6

Levels 1 and 2 are easy, because level 1 is just the root and level 2 is just its children, but after that I can't figure out how to get it to get the children of the children.

Answer 1

One way of doing so would be using a queue data structure.

Start with some queue q , and place in the index of the (unique) item whose parent is -1. Now, at each step, until q is empty,

• Perform v <- pop(q) (popping the head)
• Print out v
• For each child w of v , do push(q, v) (pushing ot the tail)

For example, here are the first steps for your case:

• Initially, q = [3] (3 is the index of the item whose parent is -1).
• We pop q , print out 3, and push 0, 2, and 4, so q = [0, 2, 4] .
• Now we pop q , print out 0, and push 1, so q = [2, 4, 1] .

Almost by definition, since q is popped from the front and added to the back, the nodes will be processed level by level.

The complexity is linear in the number of nodes.

Answer 2

You will have to perform a BFS (Breadth First Search) on the tree, while maintaining the number of nodes pushed into the next level. Outline:

q.push(root); nodesInCurrentLevel = 1; nodesInNextLevel = 0; currentLevelIndex = 1;
while q is not empty do:
  u = q.pop()
  print currentLevelIndex and u
  decrement nodesInCurrentLevel
  for every child v of u do:
    increment nodesInNextLevel
    q.push(v)
  if nodesInCurrentLevel is 0 do:
    nodesInCurrentLevel = nodesInNextLevel
    nodesInNextLevel = 0
    increment currentLevelIndex

Of course, this would print the output as Level 2:0 Level 2:2, etc. You can store current level nodes in a temporary list within the loop and print as appropriate.