[编程之美3.10]分层遍历二叉树

题目1描述：给定一棵二叉树，要求分层遍历该二叉树，即从上到下按层次访问该二叉树(每层将单独输出一行)，每层要求访问的顺序从左到右，并将节点依次编号。正确输出为：

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1

2 3

4 5 6

7 8

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#include <iostream>#include <vector>using namespace std;struct Node{Node * pLeft;Node * pRight;int value;Node(int v) : value(v), pLeft(NULL), pRight(NULL){}};voidPrintNodeByLevel(Node * root){vector<Node *>vec;intcur = 0;intlast = 1;if(root == NULL)return ;vec.push_back(root);while(cur < vec.size()){last = vec.size();while(cur < last){cout << vec[cur]->value << " ";if(vec[cur]->pLeft != NULL)vec.push_back(vec[cur]->pLeft);if(vec[cur]->pRight != NULL)vec.push_back(vec[cur]->pRight);cur++;}cout << endl;}cout << endl;}int main(){Node* root = new Node(1);    Node* tmp = new Node(2);    root->pLeft = tmp;    tmp = new Node(3);    root->pRight = tmp;    tmp = new Node(4);    root->pLeft->pLeft = tmp;    tmp = new Node(5);    root->pLeft->pRight = tmp;    tmp = new Node(6);    root->pRight->pRight = tmp;    tmp = new Node(7);    root->pLeft->pRight->pLeft = tmp;    tmp = new Node(8);    root->pLeft->pRight->pRight = tmp;PrintNodeByLevel(root);system("pause");}

扩展题1：

如果要求按深度从下到上访问图中的二叉树，每层的访问顺序仍然是从左到右，即访问顺序变为：

—————————————————————————————————————————————————————————————————————————————

7 8

4 5 6

2 3

1

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#include <iostream>#include <vector>using namespace std;struct Node{Node * pLeft;Node * pRight;int value;Node(int v) : value(v), pLeft(NULL), pRight(NULL){}};voidPrintNodeByLevel(Node * root){vector<Node *>vec;intcur = 0;intlast = 1;if(root == NULL)return ;vec.push_back(root);while(cur < vec.size()){last = vec.size();vec.push_back(NULL);while(cur < last){if(vec[cur]->pLeft != NULL)vec.push_back(vec[cur]->pLeft);if(vec[cur]->pRight != NULL)vec.push_back(vec[cur]->pRight);cur++;}cur++;}for(vector<Node*>::const_reverse_iterator iter = vec.rbegin() + 1; iter != vec.rend(); iter++){if(*iter == NULL )cout << endl;elsecout << (*iter)->value << " ";}cout << endl;}int main(){Node* root = new Node(1);    Node* tmp = new Node(2);    root->pLeft = tmp;    tmp = new Node(3);    root->pRight = tmp;    tmp = new Node(4);    root->pLeft->pLeft = tmp;    tmp = new Node(5);    root->pLeft->pRight = tmp;    tmp = new Node(6);    root->pRight->pRight = tmp;    tmp = new Node(7);    root->pLeft->pRight->pLeft = tmp;    tmp = new Node(8);    root->pLeft->pRight->pRight = tmp;PrintNodeByLevel(root);system("pause");}

扩展题2：

如果按照深度从下访问，每层的访问顺序从右到左，即访问顺序变为：

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8 7

6 5 4

3 2

1

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解法代码如下：

#include <iostream>#include <vector>using namespace std;struct Node{Node * pLeft;Node * pRight;int value;Node(int v) : value(v), pLeft(NULL), pRight(NULL){}};voidPrintNodeByLevel(Node * root){vector<Node *>vec;intcur = 0;intlast = 1;if(root == NULL)return ;vec.push_back(root);while(cur < vec.size()){last = vec.size();vec.push_back(NULL);while(cur < last){if(vec[cur]->pRight != NULL)vec.push_back(vec[cur]->pRight);if(vec[cur]->pLeft != NULL)vec.push_back(vec[cur]->pLeft);cur++;}cur++;}for(vector<Node*>::const_reverse_iterator iter = vec.rbegin() + 1; iter != vec.rend(); iter++){if(*iter == NULL )cout << endl;elsecout << (*iter)->value << " ";}cout << endl;}int main(){Node* root = new Node(1);    Node* tmp = new Node(2);    root->pLeft = tmp;    tmp = new Node(3);    root->pRight = tmp;    tmp = new Node(4);    root->pLeft->pLeft = tmp;    tmp = new Node(5);    root->pLeft->pRight = tmp;    tmp = new Node(6);    root->pRight->pRight = tmp;    tmp = new Node(7);    root->pLeft->pRight->pLeft = tmp;    tmp = new Node(8);    root->pLeft->pRight->pRight = tmp;PrintNodeByLevel(root);system("pause");}

扩展题3：(参照leetcode中的Binary Tree Zigzag Level Order Traversal)

若访问顺序为：

————————————————————————————————————————————————————————————————————————————

1

3 2

4 5 6

8 7

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#include <iostream>#include <vector>using namespace std;struct Node{Node * pLeft;Node * pRight;int value;Node(int v) : value(v), pLeft(NULL), pRight(NULL){}};voidPrintNodeByLevel(Node * root){vector<Node *>vec;vector<vector<int> > result;vector<int>val;intcur = 0;intlast = 1;if(root == NULL)return ;vec.push_back(root);bool order = false;while(cur < vec.size()){last = vec.size();while(cur < last){val.push_back(vec[cur]->value);if(vec[cur]->pLeft != NULL)vec.push_back(vec[cur]->pLeft);if(vec[cur]->pRight != NULL)vec.push_back(vec[cur]->pRight);cur++;}order = !order;if(!order)reverse(val.begin(),val.end());result.push_back(val);val.clear();}for(vector<vector<int> >::iterator iter = result.begin(); iter != result.end(); iter++){for(vector<int>::iterator cr = (*iter).begin(); cr != (*iter).end(); cr++)cout << *cr << " ";cout << endl;}}int main(){Node* root = new Node(1);    Node* tmp = new Node(2);    root->pLeft = tmp;    tmp = new Node(3);    root->pRight = tmp;    tmp = new Node(4);    root->pLeft->pLeft = tmp;    tmp = new Node(5);    root->pLeft->pRight = tmp;    tmp = new Node(6);    root->pRight->pRight = tmp;    tmp = new Node(7);    root->pLeft->pRight->pLeft = tmp;    tmp = new Node(8);    root->pLeft->pRight->pRight = tmp;PrintNodeByLevel(root);system("pause");}