二叉树的创建、遍历
二叉树的创建。这里采用最简单的情况，创建完全二叉树，用数组来保存：
1 struct TreeNode
2 {
3 int val;
4 TreeNode *left, *right;
5 TreeNode(int x): val(x), left(NULL), right(NULL) {};
6 };
7
8 void CreatTree(TreeNode *root, int idx, int A[], int n) //root结点已分配，idx为根结点的编号
9 {
10 if(2 * idx <= n) //左儿子
11 {
12 root->left = new TreeNode(A[2 * idx - 1]);
13 CreatTree(root->left, 2 * idx, A, n);
14 }
15 if(2 * idx + 1 <= n) //右儿子
16 {
17 root->right = new TreeNode(A[2 * idx]);
18 CreatTree(root->right, 2 * idx + 1, A, n);
19 }
20 }
二叉树的销毁：
1 void DestroyTree(TreeNode *root)
2 {
3 if(root == NULL)
4 return;
5 DestroyTree(root->left); //先递归删除子树
6 DestroyTree(root->right);
7 delete root;
8 }
二叉树的层次遍历：
1 //层次遍历
2 void BreadthFirstPrint(TreeNode *root) //BFS, 用队列
3 {
4 if(root == NULL)
5 return;
6 queue<TreeNode *> que;
7 que.push(root);
8
9 TreeNode *current = NULL;
10 while(!que.empty())
11 {
12 current = que.front();
13 que.pop();
14
15 cout << current->val << ' ';
16
17 if(current->left != NULL)
18 que.push(current->left);
19 if(current->right != NULL)
20 que.push(current->right);
21 }
22 }
二叉树的递归遍历：
1 //先序遍历，递归
2 void PreorderRe(TreeNode *root)
3 {
4 if(root == NULL)
5 return;
6 cout << root->val << ' ';
7 PreorderRe(root->left);
8 PreorderRe(root->right);
9 }
10
11 //中序遍历，递归
12 void InorderRe(TreeNode *root)
13 {
14 if(root == NULL)
15 return;
16 InorderRe(root->left);
17 cout << root->val << ' ';
18 InorderRe(root->right);
19 }
20
21 //后序遍历，递归版
22 void PostorderRe(TreeNode *root)
23 {
24 if(root == NULL)
25 return;
26 PostorderRe(root->left);
27 PostorderRe(root->right);
28 cout << root->val << ' ';
29 }
二叉树的非递归遍历，用栈：
1 //先序遍历，非递归，用栈；也是DFS算法
2 void PreorderStk(TreeNode *root)
3 {
4 if(root == NULL)
5 return;
6 stack<TreeNode *> stk;
7 stk.push(root);
8
9 TreeNode *current = NULL;
10 while(!stk.empty())
11 {
12 current = stk.top();
13 stk.pop();
14
15 cout << current->val << ' ';
16
17 if(current->right != NULL)
18 stk.push(current->right);
19 if(current->left != NULL)
20 stk.push(current->left);
21 }
22 }
23
24 //中序遍历，用栈，非递归
25 void InorderStk(TreeNode *root)
26 {
27 if(root == NULL)
28 return;
29 stack<TreeNode *> stk;
30 stk.push(root);
31
32 TreeNode *current = root->left;
33 while(!stk.empty() || current != NULL) //中序遍历，左子树访问完后栈为空，所以同时检测current，此时它指向右子树根
34 {
35 if(current != NULL) //向左遍历到底
36 {
37 stk.push(current);
38 current = current->left;
39 }
40 else //左子树为空，访问元素后遍历其右子树
41 {
42 cout << stk.top()->val << ' ';
43 current = stk.top()->right;
44 stk.pop();
45 }
46 }
47 }
48
49 //后序遍历，非递归版，用栈
50 void PostorderStk(TreeNode *root)
51 {
52 stack<TreeNode *> ndStk;
53 TreeNode *cur = root, *prev = NULL; //维持一个prev指针，指向上一次访问的结点
54
55 while(cur != NULL) //先向左遍历到底，入栈
56 {
57 ndStk.push(cur);
58 cur = cur->left;
59 }
60
61 while(!ndStk.empty())
62 {
63 cur = ndStk.top();
64 if(cur->right == prev) //右子树的根已访问，说明右子树访问完；或者右子树为空，则访问当前结点
65 {
66 cout << cur->val << ' ';
67 ndStk.pop();
68 prev = cur;
69 }
70 else //访问右子树
71 {
72 cur = cur->right;
73 while(cur != NULL) //与处理根节点时的情况相同，向左遍历到底入栈，prev赋空值
74 {
75 ndStk.push(cur);
76 cur = cur->left;
77 }
78 prev = NULL;
79 }
80 }//while
81 }
二叉树的非递归遍历，O(1)空间，Morris遍历：
1 //先序遍历，非递归，O(1)空间。Morris遍历
2 void PreorderMrs(TreeNode *root)
3 {
4 TreeNode *p = root;
5 while(p != NULL)
6 {
7 if(p->left != NULL) //找中序遍历的前驱结点
8 {
9 TreeNode *tmp = p->left;
10 while(tmp->right != NULL && tmp->right != p)
11 tmp = tmp->right;
12 if(tmp->right == NULL) //前驱还未做标记，说明当前结点还没访问
13 {
14 tmp->right = p; //标记前驱结点，使其右儿子为当前结点
15 cout << p->val << ' ';
16 p = p->left; //访问当前节点后访问左子树
17 }
18 else //前驱已做标记，说明当前结点与左子树已访问过
19 {
20 tmp->right = NULL;
21 p = p->right;
22 }
23 }
24 else //左子树为空，访问当前节点后访问右子树
25 {
26 cout << p->val << ' ';
27 p = p->right;
28 }
29 }
30 }
31
32 //中序遍历，非递归，O(1)空间。Morris遍历。
33 void InorderMrs(TreeNode *root)
34 {
35 TreeNode *p = root;
36 while(p != NULL)
37 {
38 if(p->left != NULL)
39 {
40 TreeNode *tmp = p->left;
41 while(tmp->right != NULL && tmp->right != p)
42 tmp = tmp->right;
43 if(tmp->right == NULL)
44 {
45 tmp->right = p;
46 p = p->left;
47 }
48 else
49 {
50 cout << p->val << ' '; //和先序Morris的唯一不同点，访问元素的位置不同。当左子树访问完之后再访问当前结点。
51 p = p->right;
52 tmp->right = NULL;
53 }
54 }
55 else
56 {
57 cout << p->val << ' ';
58 p = p->right;
59 }
60 }
61 }
62
63 //后序遍历，非递归，O(1)空间。Morris遍历。需要两个辅助函数
64 void reverse(TreeNode *start, TreeNode *stop)
65 {
66 TreeNode *prev = start, *cur = start->right, *next;
67 while(prev != stop)
68 {
69 next = cur->right;
70 cur->right = prev;
71 prev = cur;
72 cur = next;
73 }
74 }
75 void printReverseList(TreeNode *start, TreeNode *stop)
76 {
77 TreeNode *p = stop;
78 reverse(start, stop);
79 while(true)
80 {
81 cout << p->val << ' ';
82 if(p == start)
83 break;
84 p = p->right;
85 }
86 reverse(stop, start);
87 }
88 void PostorderMrs(TreeNode *root)
89 {
90 TreeNode dummy(-1), *p = &dummy;
91 dummy.left = root; //新建一个结点，把root作为左儿子，方便统一处理
92
93 while(p != NULL)
94 {
95 if(p->left != NULL) //找中序遍历的前驱结点
96 {
97 TreeNode *tmp = p->left;
98 while(tmp->right != NULL && tmp->right != p)
99 tmp = tmp->right;
100 if(tmp->right == NULL)
101 {
102 tmp->right = p;
103 p = p->left;
104 }
105 else
106 {
107 printReverseList(p->left, tmp); //与中序遍历的不同点，逆序输出当前结点的左儿子到前驱之间的路径
108
109 tmp->right = NULL;
110 p = p->right;
111 }
112 }
113 else
114 {
115 p = p->right;
116 }
117 }//while
118 }
二叉树按列打印，如下所示：
输出为：
4
2
1 5 6
3
7
思路：如果根节点对应行号为0，则2节点行号为-1，4节点行号为-2；同理3节点行号为1，7节点行号为2。即左儿子行号为父亲行号-1，右儿子则+1。
1 void ColFirstPrint(TreeNode *root)
2 {
3 if(root == NULL)
4 return;
5 int minCol = 0, maxCol = 0;
6
7 TreeNode *pt = root;
8 while(pt->left != NULL) //先求出最左节点的行号
9 {
10 pt = pt->left;
11 --minCol;
12 }
13
14 pt = root;
15 while(pt->right != NULL) //再求出最右节点的行号
16 {
17 pt = pt->right;
18 ++maxCol;
19 }
20
21 vector<vector<TreeNode*> > lines(maxCol - minCol + 1); //总的行号可计算出来
22
23 preOrder(root, -minCol, lines); //先序遍历，根节点行号为最左节点行号的绝对值
24
25 for(int i = 0; i < lines.size(); ++i)
26 {
27 for(int j = 0; j < lines[i].size(); ++j)
28 cout << lines[i][j]->val << ' ';
29 cout << endl;
30 }
31 }
32
33 void preOrder(TreeNode *root, int col, vector<vector<TreeNode*> > &lines)
34 {
35 if(root != NULL)
36 {
37 lines[col].push_back(root);
38 if(root->left != NULL)
39 preOrder(root->left, col - 1, lines);
40 if(root->right != NULL)
41 preOrder(root->right, col + 1, lines);
42 }
43 }
