获取树中节点的祖先

Question

My tree looks something like this. I need to get all ancestors of a given node. A node can have multiple children and each child can in turn be a parent.

My code looks like this, this keeps returning undefined. Please advise.

  getAncestors(nodeId: number, ancestors: number[] = []): number[] {
    if (this.nodeId === nodeId) {
      return ancestors;
    }

    if (this.children.length > 0) {
      ancestors.push(this.nodeId);
    }

    for (const child of this.children) {
      child.getAncestors(nodeId,  ancestors);
    }

    return ancestors;
  }

Answer 1

Just reading this line should indicate a major red flag

child.getAncestors(nodeId,  ancestors);

If every time someone asks who all my ancestors are, I have to first ask my children who their parents are, at best my procedure is incredibly inefficient, and at worst this will encounter an infinite loop. We should not call children in this function at all.

A correct solution would be dependent on the rest of this node class. For example, if every node has access to the root, we could search from the root until we find the original node and track all the nodes we traveled along the way. This could get messy.

An easier solution would be to track all ancestors as each node is being created. Consider the case of a file system, where the absolute path is required to access any file. You do need to update this list of ancestors if you predict you will rebalance and move nodes around often.

If nodes had access to their parents this can be done with a simple while loop and a pointer

ancestors = [];
nodePtr = this;
while (nodePtr !== root) {
    nodePtr = nodePtr.parent();
    ancestors.push(nodePtr);
}
nodePtr.push(root);