二叉树的建立和遍历

Posted 2022-02-20 blue-lin

Date：2019-06-28 13:51:23

二叉树的建立

• 注意一下中序和层序建树

 1 
 2 //给出当前先序序列PreL, PreL+1, ..., Pre(L+numLeft), Pre(L+numLeft+1), ..., PreR
 3 //给出当前中序序列InL, InL+1, ..., In(K-1), InK, In(k+1), ..., InR
 4 BiTnode* CreatePreIn(int preL, int preR, int inL, int inR)
 5 
 6     if(preL > preR)
 7         return NULL;    //先序序列长度小于0
 8 
 9     BiTnode *root = new BiTnode;
10     root->data = pre[preL];
11     int k;
12     for(int k=inL; k<=inR; k++)
13         if(in[k] == pre[preL])  //在中序序列中找到根节点
14             break;
15     int numLeft = k - inL;      //由中序序列得到左子树结点的个数
16 
17     //左子树的先序区间为[preL+1, preL+numLeft], 左子树的中序区间为[inL, k-1];
18     root->lchild = CreatePreIn(preL+1, preL+numLeft, inL, k-1);
19     //右子树的先序区间为[preL+numLeft+1, preR], 右子树的中序区间为[k+1, inR];
20     root->rchild = CreatePreIn(preL+numLeft+1, preR, k+1, inR);
21 
22     return root;
23 
24 
25 //给出当前后序序列PostL, PostL+1, ..., Post(L+numLeft-1), Post(L+numLeft), ..., Pre(R-1), PreR
26 //给出当前中序序列InL, InL+1, ..., In(k-1), Ink, In(k+1), ..., InR
27 BiTnode* CreatePostIn(int postL, int postR, int InL, int inR)
28 
29     if(postL > postR)
30         return NULL;
31 
32     BiTnode *root = new BiTnode;
33     root->data = post[postR];
34     int k;
35     for(int k=inL; k<=inR; k++)
36         if(In[k] == post[postR])
37             break;
38     int numleft = k - inL;
39 
40     root->lchild = CreatePostIn(postL, postL+numLeft-1, inL, k-1);
41     root->rchild = CreatePostIn(postL+numLeft, postR-1, k+1, inR);
42 
43     return root;
44 
45 
46 //给出当前中序序列InL, ..., In(k-1), Ink, In(k+1), ..., InR
47 //给出当前层次序列layerL, ..., layerR
48 BiTnode* CreateLayerIn(int inL, int inR, int layerL, int layerR)
49 
50     if(inL > inR)
51         return NULL;
52     BiTnode *root = new BiTnode;
53 
54     int i, j;
55     for(int i=layerL; i<=layerR; i++)
56     
57         int x=0;
58         for(int j=inL; j<=inR; j++)
59         
60             if(layer[i] = in[j])
61             
62                 b = 1;
63                 root->data = in[j];
64                 break;
65             
66         
67         if(b)   break;
68     
69     root->lchild = CreateLayerIn(inL, j-1, layerL, layerR);
70     root->rchild = CreateLayerIn(j+1, inR, layerL, layerR);
71 
72     return root;
73 

二叉树的遍历

  1 //先序
  2 void PreOrder(BiTnode* T)
  3 
  4     if(T == NULL)
  5         return;
  6     //visit(T);
  7     PreOrder(T->lchild);
  8     PreOrder(T->rchild);
  9 
 10 
 11 //先序非递归
 12 #include <stack>
 13 using namespace std;
 14 void PreOrder(BiTnode* T)
 15 
 16     stack<BiTnode*> s;
 17     BiTnode *p = T;
 18     while(p || !s.empty())
 19     
 20         if(p)
 21         
 22             //visit(p);
 23             s.push(p);
 24             p = p->lchild;
 25         
 26         else
 27         
 28             p = s.top();
 29             s.pop();
 30             p = p->rchild;
 31         
 32     
 33 
 34 
 35 //中序
 36 void InOrder(BiTnode* T)
 37 
 38     if(T == NULL)
 39         return
 40     InOrder(T->lchild);
 41     //visit(T);
 42     InOrder(T->rchild);
 43 
 44 
 45 //中序非递归
 46 #include <stack>
 47 using namespace std;
 48 void InOrder2(BiTnode* T)
 49 
 50     stack<BiTnode*> s;
 51     BiTnode *p = T;
 52     while(p || !s.empty())
 53     
 54         if(p)
 55         
 56             s.push(p);
 57             p = p->lchild;
 58         
 59 else
 60 
 61 p = s.top();
 62 //visit(p);
 63 s.pop();
 64 p = p->rchild;
 65 
 66 
 67 
 68 
 69 //后序
 70 void PostOrder(BiTnode* T)
 71 
 72     if(T == NULL)
 73         return;
 74     PostOrder(T->lchild);
 75     PostOrder(T->rchild);
 76     //visit(T);
 77 
 78 
 79 //后序非递归
 80 void PostOrder(BiTnode* T)
 81 
 82     stack<BiTnode*> s;
 83     BiTnode *p, *r=NULL;
 84     while(p || !s.empty())
 85     
 86         if(p)
 87         
 88             s.push(p);
 89             p = p->lchild;
 90         
 91         else
 92         
 93             p = s.top();
 94             if(p->rchild && p->rchild!=r)
 95             
 96                 p = p->rchild;
 97                 s.push(p);
 98                 p = p->lchild;
 99             
100             else
101                //p出栈时，栈内为根节点到p的路径，以此可以求解公共结点等
102                 //visit(p);
103                 s.pop();
104                 r = p;
105                 p = NULL;
106             
107         
108     
109 
110 
111 //层次
112 #include <queue>
113 using namespace std;
114 void LevelOrder(BiTnode* T)
115 
116     queue<BiTnode*> q;      //队列中存放BiTnode变量的地址，这样就可以通过访问地址去修改原元素
117     T->level = 1;           //记录结点深度
118     q.push(T);
119     while(!q.empty())
120     
121         BiTnode *p = q.front();
122         q.pop();
123         //visit(p);
124         if(p->lchild)
125         
126             p->lchild->level = p->level + 1;       //计算各结点深度
127             q.push(p->lchild);
128         
129         if(p->rchild)
130         
131             p->rchild->level = p->level + 1;
132             q.push(p->rchild);
133         
134     
135 

多叉树的静态表示

 1 //存储结构：孩子表示法
 2 #include <vector>
 3 using namespace std;
 4 const int M = 1e3;
 5 struct node
 6 
 7     int layer;  //结点深度
 8     int data;
 9     vector<int> children;
10 tree[M];
11 
12 //若结点不涉及数据域，可以简化结点结构
13 vector<int> child[M];
14 
15 //插入结点
16 int index=0;
17 int NewNode(int x)
18 
19     tree[index].data = x;
20     tree[index].children.clear();
21     return index++;
22 
23 
24 //遍历
25 void PreOrder(int root)
26 
27     printf("%d ", tree[root].data);     //先根遍历
28     for(int i=0; i<tree[root].children.size(); i++)
29         PreOrder(tree[root].children[i]);
30     Printf("%d ", tree[root].data);     //后根遍历
31 
32 
33 //层序遍历
34 #include <queue>
35 using namespace std;
36 void LayerOrder(int root)
37 
38     queue<int> q;
39     tree[root].layer = 0;
40     q.push(root);
41     while(!q.empty())
42     
43         root = q.front();
44         q.pop();
45         printf("%d ", tree[root].data);
46         for(int i=0; i<tree[root].children.size(); i++)
47         
48             int child = tree[root].children[i];
49             tree[child].layer = tree[root].layer+1;
50             q.push(child);
51         
52     
53