Convert an ordered array to a binary tree

This question is a bit ambiguous, just to convert it into a binary tree. This does not mean a balanced binary tree or a binary sorting tree.

If you can use an array as the output for any binary tree, you can use the pre-order, middle-order, or subsequent binary tree construction. There is a lot of code for building a binary tree on the Internet.

The following code is a binary sorting tree of resumes.

First, use the intermediate element of the array as the root to establish a binary tree, and then recursively return the left and right subtree of the resume.

# Include <iostream> # include <queue> # include <windows. h> using namespace STD; typedef struct node {int data; struct node * left; struct node * right;} node, * bitree; void buildtree (bitree & T, int A [], int begin, int end) // create a binary sorting tree {If (begin> end) // recursive exit return; int mid = (begin + end) /2; // subscript of the intermediate element if (t = NULL) // apply for space for the root node of the current tree {T = (node *) malloc (sizeof (node); t-> DATA = A [Mid]; // value assignment t-> left = NULL; // left and right subtree must also be set to null T-> righ T = NULL;} cout <A [Mid] <""; // The following two actions show the creation process sleep (1000); buildtree (t-> left, a, begin, mid-1); // recursive resume left subtree buildtree (t-> right, A, Mid + 1, end ); // recursively create the right subtree} void travel (bitree t) {If (T! = NULL) {travel (t-> left); cout <t-> data <""; // traverse travel (t-> right) in the middle order );}} int main () {int A [] = {1, 3, 4, 5, 6, 7, 8, 9}; bitree root = NULL; int begin = 0, end = 8; buildtree (root, a, begin, end); cout <Endl; travel (Root); getchar (); Return 0 ;}