How to infer that a binary tree is a Complete Binary Tree

Yan Weimin's statement in this textbook: a binary tree with a depth of K and a number of nodes of 2 ^ k-1 is a full binary tree. This concept is very understandable,

It is a tree with a depth of K and no space available.

First, the full binary tree is numbered in the order of breadth first traversal (from left to right.

A k-tree with N nodes in depth also numbers the tree. If all the numbers correspond to full-tree trees, the tree is a Complete Binary Tree.

 

A random binary tree can be combined into a full binary tree. In this way, there will be many holes in the middle. In the case of breadth-first traversal, assume that it is a full binary tree or a full binary tree. These holes are at the end of the breadth-first traversal. Therefore, when we traverse to holes, the entire binary tree has been traversed. Assume that it is a non-Complete Binary Tree,

When we traverse the holes, we will find that there are no traversal values behind the holes. In this way, we only need to infer whether a Complete Binary Tree is based on whether to traverse to a hole or whether the whole tree traversal is over.

The algorithm is as follows:

Bool is_complete (TREE * root) {queue Q; tree * PTR; // performs breadth-first traversal (hierarchical traversal) and puts null nodes into the queue Q. push (Root); While (PTR = Q. pop ())! = NULL) {q. Push (PTR-> left); q. Push (PTR-> right);} // infer whether there are still nodes not asked by the operator while (! Q. is_empty () {PTR = Q. Pop (); // if a non-empty node is not found, the tree has holes. If (null! = PTR) {return false ;}} return true ;}

How to infer that a binary tree is a Complete Binary Tree