forked from H-K-ai/data-structure-and-algorithm
-
Notifications
You must be signed in to change notification settings - Fork 0
Expand file tree
/
Copy pathbinary_tree.cpp
More file actions
385 lines (309 loc) · 8.94 KB
/
Copy pathbinary_tree.cpp
File metadata and controls
385 lines (309 loc) · 8.94 KB
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
173
174
175
176
177
178
179
180
181
182
183
184
185
186
187
188
189
190
191
192
193
194
195
196
197
198
199
200
201
202
203
204
205
206
207
208
209
210
211
212
213
214
215
216
217
218
219
220
221
222
223
224
225
226
227
228
229
230
231
232
233
234
235
236
237
238
239
240
241
242
243
244
245
246
247
248
249
250
251
252
253
254
255
256
257
258
259
260
261
262
263
264
265
266
267
268
269
270
271
272
273
274
275
276
277
278
279
280
281
282
283
284
285
286
287
288
289
290
291
292
293
294
295
296
297
298
299
300
301
302
303
304
305
306
307
308
309
310
311
312
313
314
315
316
317
318
319
320
321
322
323
324
325
326
327
328
329
330
331
332
333
334
335
336
337
338
339
340
341
342
343
344
345
346
347
348
349
350
351
352
353
354
355
356
357
358
359
360
361
362
363
364
365
366
367
368
369
370
371
372
373
374
375
376
377
378
379
380
381
382
383
384
385
/**
* @Author: Chacha
* @Date: 2018-11-24 13:31:24
* @Last Modified by: Chacha
* @Last Modified time: 2018-12-19 22:55:04
* @Source: https://leetcode.com/explore/learn/card/data-structure-tree
*/
#include <iostream>
#include <vector>
#include <stack>
#include <deque>
using namespace std;
/**
* Definition for a binary tree node.
*/
struct TreeNode {
int val;
int isFirst;
TreeNode *left;
TreeNode *right;
TreeNode(int x) : val(x), left(NULL), right(NULL) {}
};
/**
* Pre-order Traversal
* Pre-order traversal is to visit the root first. Then traverse the left subtree.
* Finally, traverse the right subtree. Here is an example:
* https://leetcode.com/explore/learn/card/data-structure-tree/134/traverse-a-tree/992/#pre-order-traversal
*/
// Recursive Way
void preorderTraversal(TreeNode* root) {
if (root == NULL) {
return;
}
cout << root->val;
preorderTraversal(root->left);
preorderTraversal(root->right);
}
// Divide & Conquer
vector<int> preorderTraversalDivideConquer(TreeNode* root) {
vector<int> result;
if (root == NULL) {
return result;
}
vector<int> left = preorderTraversalDivideConquer(root->left);
vector<int> right = preorderTraversalDivideConquer(root->right);
result.push_back(root->val);
result.insert(result.end(), left.begin(), left.end());
result.insert(result.end(), right.begin(), right.end());
return result;
}
// Iterative Way 1
void iterativePreOrderTraversal1(TreeNode* root) {
if(root == NULL) {
cout << "The tree is NULL..." << endl;
}
stack<TreeNode*> nStack;
TreeNode* node = root;
// Start to traversal
while(node != NULL || nStack.empty() != true) {
// Output current subtree root node, and iterate to the last left node
while(node != NULL){
cout << node->val;
nStack.push(node);
node = node->left;
}
/**
* While is end, and the top of the stack is the last left node,
* and then start to pop the stack and output the right node.
*/
if (nStack.empty() != true) {
node = nStack.top();
nStack.pop();
node = node->right;
}
}
}
// Iterative Way 2
void iterativePreOrderTraversal2(TreeNode* root) {
if(root == NULL) {
cout << "The tree is NULL..." << endl;
}
stack<TreeNode*> nStack;
nStack.push(root);
TreeNode *node = root;
while(nStack.empty() != true) {
node = nStack.top();
nStack.pop();
cout << node->val;
if (node->right != NULL) {
nStack.push(node->right);
}
if (node->left != NULL) {
nStack.push(node->left);
}
}
}
/**
* In-order Traversal
* In-order traversal is to traverse the left subtree first.
* Then visit the root. Finally, traverse the right subtree.
* Here is an example:
* https://leetcode.com/explore/learn/card/data-structure-tree/134/traverse-a-tree/992/#in-order-traversal
*/
// Recursive Way
void inOrderTraversal(TreeNode* root) {
if (root == NULL) {
return;
}
inOrderTraversal(root->left);
cout << root->val;
inOrderTraversal(root->right);
}
// Divide & Conquer
vector<int> inOrderTraversalDivideConquer(TreeNode* root) {
vector<int> result;
if (root == NULL) {
return result;
}
vector<int> left = inOrderTraversalDivideConquer(root->left);
vector<int> right = inOrderTraversalDivideConquer(root->right);
result.insert(result.end(), left.begin(), left.end());
result.push_back(root->val);
result.insert(result.end(), right.begin(), right.end());
return result;
}
// Iterative Way
void iterativeInOrderTraversal(TreeNode* root) {
if (root == NULL) {
cout << "The tree is NULL..." << endl;
}
stack<TreeNode*> nStack;
TreeNode* node = root;
while(node != NULL && nStack.empty() != true) {
while(node != NULL){
nStack.push(node);
node = node->left;
}
if (nStack.empty() != true) {
node = nStack.top();
cout << node->val;
nStack.pop();
node = node->right;
}
}
}
/**
* Post-order Traversal
* Post-order traversal is to traverse the left subtree first.
* Then traverse the right subtree. Finally, visit the root.
* Here is an animation to help you understand post-order traversal:
* https://leetcode.com/explore/learn/card/data-structure-tree/134/traverse-a-tree/992/#post-order-traversal
*/
// Recursive Way
void postOrderTraversal(TreeNode* root) {
if (root == NULL) {
return;
}
postOrderTraversal(root->left);
postOrderTraversal(root->right);
cout << root->val;
}
// Divide & Conquer
vector<int> postOrderTraversalDivideConquer(TreeNode* root) {
vector<int> result;
if (root == NULL) {
return result;
}
vector<int> left = postOrderTraversalDivideConquer(root->left);
vector<int> right = postOrderTraversalDivideConquer(root->right);
result.insert(result.end(), left.begin(), left.end());
result.insert(result.end(), right.begin(), right.end());
result.push_back(root->val);
return result;
}
// Iterative Way 1
void iterativePostOrderTraversal1(TreeNode* root) {
if (root == NULL) {
cout << "The tree is NULL..." << endl;
}
stack<TreeNode*> nStack;
TreeNode* node = root;
while(root != NULL && nStack.empty() != true) {
while(node != NULL){
node->isFirst = 1; // When the node be accessed at first
nStack.push(node);
node = node->left;
}
if (nStack.empty() != true) {
node = nStack.top();
nStack.pop();
if (node->isFirst == 1) {
node->isFirst++;
nStack.push(node);
node = node->right;
} else if (node->isFirst == 2) {
cout << node->val;
node = NULL;
}
}
}
}
// Iterative Way 2
void iterativePostOrderTraversal2(TreeNode* root) {
if (root == NULL) {
cout << "The tree is NULL..." << endl;
}
stack<TreeNode*> nStack;
TreeNode* cur; // Current node
TreeNode* prev = NULL; // Previous node
nStack.push(root);
while(nStack.empty() != true){
cur = nStack.top();
/**
* 1. The left and right child are NULL
* 2. The children are output
*/
if ((cur->left == NULL && cur->right == NULL) ||
(prev != NULL && ((prev == cur->left) || prev == cur->right))
) {
cout << cur->val;
nStack.pop();
prev = cur;
} else {
if (cur->right != NULL) {
nStack.push(cur->right);
}
if (cur->left != NULL) {
nStack.push(cur->left);
}
}
}
}
/**
* Binary Tree Level Order Traversal
* Source: https://shimo.im/docs/MAbjWlrqWqU1f72m/
*/
// Recursive Way 1
int PrintLevel1(TreeNode* root, int n, int level) {
if (root == NULL || level < 0) {
return 0;
} else if (level == n) {
cout << root->val;
return 1;
} else {
return PrintLevel1(root->left, n, level + 1) + PrintLevel1(root->right, n, level + 1);
}
}
// Recursive Way 2
int PrintLevel2(TreeNode* root, int n, int level) {
if (root == NULL || level < 0) {
return 0;
} else if (level == n) {
cout << root->val;
return 1;
} else {
return PrintLevel2(root->left, n, level - 1) + PrintLevel2(root->right, n, level - 1);
}
}
/**
* Iterative Way 1
* Use two queues, one queue for current level nodes and another for next level nodes
*/
void LevelOrderDev(TreeNode* root) {
deque<TreeNode *> qFirst, qSecond;
qFirst.push_back(root);
while(qFirst.empty() != true){
while(qFirst.empty() != true) {
TreeNode* temp = qFirst.front();
qFirst.pop_front();
cout << temp->val;
if (temp->left != NULL) {
qSecond.push_back(temp->left);
}
if (temp->right != NULL) {
qSecond.push_back(temp->right);
}
}
cout << endl;
qFirst.swap(qSecond);
}
}
/**
* Iterative Way 1
* Use two pointers, one pointer for current level nodes and another for next level nodes
*/
void LevelOrderUsePoint(TreeNode* root) {
vector<TreeNode*> vec;
vec.push_back(root);
int cur = 0;
int end = 1;
while(cur < vec.size()){
end = vec.size();
while(cur < end){
cout << vec[cur]->val;
if (vec[cur]->left != NULL) {
vec.push_back(vec[cur]->left);
}
if (vec[cur]->right != NULL) {
vec.push_back(vec[cur]->right);
}
cur++;
}
cout << endl;
}
}
int main() {
/* code */
return 0;
}