Question

I’m deciding how to store data and how to draw a tree graph. Assuming I want a minimum space M between two elements, I was thinking I could traverse the entire tree structure from the top to the bottom in breath-first search.

If there’s just one element below the current one, it will be drawn with the same X coordinate as his father. If there are two elements, they will be drawn one at -M/2 and the other at +M/2 with respect to their father X coordinate. And so on..

The problem is: what if an element like C (see diagram below) has a great number of children?? I should restructure the entire tree since I should move the element D to the left and make space for all the E-F children of C. Moving D to the left will get the tree crooked and I will need to move B too. Moving B to the left would alter the tree’s symmetry so I’ll need to move C too and so on..

How can I draw a perfectly symmetric tree whose elements may have a large number of children?

Answer 1

Do it the other way up: compute each node’s horizontal position from those of its children after they’ve been computed. Something like this (WARNING: completely untested code; may consist entirely of bugs):

void Node::place_self(coord_t x0, coord_t y0) {
  this->y0 = y0; this->y1 = y0+height;
  if (!left && !right) {
    // This is a leaf. Put its top left corner at (x0,y0).
    this->x0 = x0; this->y0 = y0;
    this->subtree_x1 = x0+width;
  }
  else if (!left || !right) {
    // Only one child. Put this immediately above it.
    Node * child = left ? left : right;
    child->place_self(x0,y0+height+gap);
    coord_t xc = child->x0 + child->width/2;
    this->x0 = xc-width/2;
    this->subtree_x1 = max(this->x0+width, child->subtree_x1);
  }
  else {
    // Two children. Put this above their midline.
    left->place_self(x0, y0+height+gap);
    right->place_self(left->subtree_x1+gap, y0+height+gap);
    coord_t xc = (x0 + right->subtree_x1)/2;
    this->x0 = xc-width/2;
    this->subtree_x1 = max(this->x0+width, right->subtree_x1);
  }
}