Binary Tree topic

[Topic 5] Create a binary tree using the result of forward and middle traversal in <Binary Tree C ++ implementation> Histogram }. First look at the first element 1 in the previous order, representing the root of the binary tree to be constructed. Search for 1 in the sequence list. After finding 1, you can judge that the left of 1 is the left subtree of the root, on the right is the right subtree of the root, so that the rest of the middle sequence is divided into {, 2,} and {,}, and the corresponding front sequence is divided into {2, 4, 7,} and {3, 5, 6, 8}; then recursively, scale down along the forward sequence, for {4, 7, 2,} and {2, 4, 7 ,} perform the same operation. The recursive boundary is that there is only one element left in the pre-order sequence and the middle sequence, and the elements are equal. In this way, the left-side left subtree has been constructed, and then the last {5, 3, 8, 6} and {3, 5, 6, 8} Do recursion, it is not difficult. The Code is as follows. Code 5]

# Ifndef null # define null 0 # endifstruct limit {int m_value; binary_tree_node * m_left; Limit * m_right;}; Limit * construct_core (int * start_preorder, int * end_preorder, int * start_inorder, int * end_inorder); binary_tree_node * construct_binary_tree (int * pre_order, int * in_order, int Len) {If (pre_order = NULL | in_order = NULL | Len <1) return NULL; return construct_core (pre_o Rder, pre_order + len-1, in_order, in_order + len-1);} binary_tree_node * construct_core (int * start_preorder, int * end_preorder, int * start_inorder, int * end_inorder) {int root_value = start_preorder [0]; binary_tree_node * node = new binary_tree_node; node-> m_value = root_value; node-> m_left = node-> m_right = NULL; if (start_preorder = end_preorder & start_inorder = end_inorder & * start_preorder = * Start_inorder) return node; // recursive boundary int * in_start = start_inorder; while (in_start <= end_inorder & * in_start! = Root_value) ++ in_start; If (in_start = end_inorder & * in_start! = Root_value) return NULL; // This array is unreasonable. The value in the previous order does not contain int left_len = in_start-start_inorder; int * left_preorder_end = start_preorder + left_len; If (left_len> 0) {// construct the left subtree node-> m_left = construct_core (start_preorder + 1, left_preorder_end, start_inorder, in_start-1);} If (left_len <end_preorder-start_preorder) {// construct the right subtree node-> m_right = construct_core (left_preorder_end + 1, end_preorder, in_start + 1, end_inorder );} Return node ;}# include <iostream >#include <queue> using namespace STD; void level_visit (binary_tree_node * root) {If (root = NULL) return; queue <binary_tree_node *> m_queue; m_queue.push (Root); While (! M_queue.empty () {binary_tree_node * B _node = m_queue.front (); m_queue.pop (); cout <B _node-> m_value <""; if (B _node-> m_left! = NULL) m_queue.push (B _node-> m_left); If (B _node-> m_right! = NULL) m_queue.push (B _node-> m_right );}}

[Test 5] The last function above is used to print out Binary Trees and traverse them in layers.

int main(){int pre[] = {1,2,4,7,3,5,6,8};int in[] = {4,7,2,1,5,3,8,6};binary_tree_node* root = construct_binary_tree(pre, in, sizeof(pre)/sizeof(pre[0]));level_visit(root);}

[Question 18] tree sub-structure [Question 19] binary tree image [question 24] Binary Search Tree post-order traversal sequence [Question 27] Binary Search Tree and two-way linked list

(To be continued)