Algorithm learning-clue Binary Tree

Algorithm learning-clue Binary Tree
Clue Binary Tree

A clue Binary Tree adds something more to a common binary tree. What is more? A chain structure is a binary tree. Explanation: Generally, in our binary tree, leaf nodes do not have children, so pointing to null meansNULLIn the clue binary tree, the left and right children of the leaf node point to their own pioneers and successors respectively. Which node is the frontend and successor? It is the previous node and the next node in the tree traversal process. Therefore, the first traversal node has no precursor, And the last node has no successor. Generally, this is a binary tree of the central direction. Of course, there are also Binary Trees of the first direction and the binary tree of the latter direction.

            []a[]            /   \         []b[] []c[]         /   \         []d[] []e[]

The above is a binary tree. I have added a label on both sides of each node.[]I haven't added any content to this label yet. There are only two values in it. One is0One is1When a node has children0When there are no children1.
Therefore, the structure of the node is:

typedef struct TreeNode* node;struct TreeNode{    node lchild;    int ltag;    int rtag;    node rchild;    int data;};

When0Point to the child node1Point to the front or back.
The code is attached below.

Code Implementation

The Code is as follows:

node Pre = NULL;void InOrderTree(node n){    if (n == NULL) {    }else{        //        Pre = n;        if (n->lchild != NULL) {            n->ltag = 0;            InOrderTree(n->lchild);        }else{            n->ltag = 1;            n->lchild = Pre;        }        if (Pre!=NULL && Pre->rchild == NULL) {            Pre->rtag = 1;            Pre->rchild = n;        }        Pre = n;        if (n->rchild != NULL) {            n->rtag = 0;            InOrderTree(n->rchild);        }else{            n->rtag = 1;        }    }}

Test code (main ):

int main(int argc, const char * argv[]) {    node head = (node)malloc(sizeof(struct TreeNode));    head->data = 1;    node node1 = (node)malloc(sizeof(struct TreeNode));    node node2 = (node)malloc(sizeof(struct TreeNode));    node node3 = (node)malloc(sizeof(struct TreeNode));    node node4 = (node)malloc(sizeof(struct TreeNode));    node1->data = 2;    node2->data = 3;    node3->data = 4;    node4->data = 5;    head->lchild = node1;    head->rchild = node2;    node1->lchild = node3;    node1->rchild = node4;    node2->lchild = NULL;    node2->rchild = NULL;    node3->lchild = NULL;    node3->rchild = NULL;    node4->lchild = NULL;    node4->rchild = NULL;    InOrderTree(head);    return 0;}