我正在尝试构建一种特定类型的树,其中每个内部节点的数据都依赖于树的叶子。我有一个预订数组pre[] = { 0,0,0,1,2,3,0,4,5}
,其中每个" 0"表示内部节点,其他任何表示叶子。如果我要构建这个树,它看起来像这样:
0
/ \
0 0
/ \ / \
0 3 4 5
/ \
1 2
我遇到了递归调用的问题,特别是在while循环中。在我完成创建带有数据" 2"的节点之后,在我返回根之后,我就离开了它的正上方的内部节点。这个内部节点将继续在while循环中运行并弄乱我的树。如果我决定在条件语句末尾的while循环中放置一个返回根,那么构造将在填充树的左侧后停止。我该如何解决这个问题?
1 #include <stdio.h>
2 #include <stdlib.h>
3 #include <limits.h>
4
5 struct node
6 {
7 int data;
8 int leafNode;
9 struct node *left;
10 struct node *right;
11 };
12
14 struct node* newNode (int data){
15 struct node* temp = (struct node *) malloc(sizeof(struct node));
16 temp->data =data;
17 temp->left = temp->right= NULL;
18 return temp;
19 }
22 struct node* _constructTree( int pre[], int * preIndex, int low, int high, int size, int leaf){
23 if (*preIndex >= size || low > high){
24 return NULL;}
25
26 printf("We are about to create this node: %d\n", pre[*preIndex]);
27
28 struct node* root = newNode(pre[*preIndex]);
29 *preIndex = *preIndex + 1;
30 if(leaf){
31 printf("Returning this leaf node: %d\n", pre[*preIndex - 1]);
32 return root;}
33
34
35 if(low == high){
36 return root;}
37
38 while(low<high){
39 if(pre[*preIndex] == 0){
40 root->left = _constructTree(pre, preIndex, *preIndex, high, size, 0);}
41
42 if(root->left == NULL){
43 root->left = _constructTree(pre, preIndex, *preIndex, high, size, 1);
44 return root;
45 }
46 if(pre[*preIndex] != 0){
47 root->right = _constructTree(pre, preIndex, *preIndex, high, size, 1);
48 return root;
49 }
50 else{
51 root->right = _constructTree(pre,preIndex, *preIndex, high, size, 0);}
52 }
53 return root;
54 }
55 }
56
57 struct node * constructTree(int pre[], int size){
58 int preIndex = 0;
59 return _constructTree(pre, &preIndex, 0, size -1, size, 0);
60 }
61
答案 0 :(得分:0)
这样的事情不会起作用(略有概括,但希望它显示出这个想法)?:
construct_tree(int pre[], int* index){
root=pre[*index];
*index++;
if(root>0)return root;
left=construct_tree(pre,index);
right=construct_tree(pre,index);
return root;
}