二叉树的层序遍历

/* * Copyright (c) 2011 alexingcool. All Rights Reserved. */ #include <iostream>using namespace std;struct Node {Node(int i = 0, Node *l = NULL, Node *r = NULL) : data(i), left(l), right(r) {}int data;Node *left;Node *right;};Node* Construct() {Node *node4 = new Node(7, NULL, new Node(3));Node *node3 = new Node(4);Node *node2 = new Node(12);Node *node1 = new Node(5, node3, node4);Node *root = new Node(10, node1, node2);return root;}int PrintByLevel(Node *root, int level) {if(root == NULL)return 0;if(level == 1) {cout << root->data << " ";return 1;}return PrintByLevel(root->left, level - 1) + PrintByLevel(root->right, level - 1);}int GetTreeHeight(Node *root) {if(root == NULL)return 0;return max(GetTreeHeight(root->left) + 1, GetTreeHeight(root->right) + 1);}void main() {Node *root = Construct();int height = GetTreeHeight(root);for(int i = 1; i <= height; i++) {PrintByLevel(root, i);cout << endl;}}

 

关于层序遍历的STL代码如下，注意reserve的妙用，否则会因为vector容量增长导致跌代器失效：

void Print2(Node *head){vector<Node*> nvec;nvec.reserve(100);nvec.push_back(head);while(!nvec.empty()) {int size = nvec.size();for(vector<Node*>::const_iterator iter = nvec.begin(); iter != nvec.end(); ++iter) {cout << (*iter)->data << " ";}cout << endl;for(vector<Node*>::const_iterator iter = nvec.begin(); iter != nvec.begin() + size; ++iter) {if((*iter)->left)nvec.push_back((*iter)->left);if((*iter)->right)nvec.push_back((*iter)->right);}nvec.erase(nvec.begin(), nvec.begin() + size);}}

 

下面代码不知为何总报错，伤感了，眼看找工作高峰来临，时间紧迫，扔这了。。。

void printLevel(Node *root){if(root == NULL)return;vector<Node *> nvec;nvec.reserve(100);nvec.push_back(root);vector<Node *>::iterator startIter = nvec.begin();vector<Node *>::iterator lastIter = nvec.end();while(startIter != lastIter) {while(startIter != lastIter) {cout << (*startIter)->data << " ";Node *left = (*startIter)->left;Node *right = (*startIter)->right;if(left)nvec.push_back(left);if(right)nvec.push_back(right);++startIter;}cout << endl;lastIter = nvec.end();}}

 再次写二叉树的层序遍历，现在的代码可以说是简洁高效，窝稀饭

void printLevel(Node *root){if (root == NULL)return;vector<Node *> nvec;nvec.push_back(root);int start = 0;int end = 1;int pos = end;while (start != end){Node *node = nvec[start];cout << node->item << " ";start++;if (node->left){nvec.push_back(node->left);end++;}if (node->right){nvec.push_back(node->right);end++;}if (start == pos){pos = end;cout << endl;}}}

关于拓展问题：

访问顺序变为：

78

456

23

1

访问顺序变为：

87

654

32

1

用栈很容易实现~

void putNode2Stack(Node *root, stack<Node *> &snode){if (root == NULL)return;vector<Node *> nvec;nvec.push_back(root);snode.push(root);snode.push(NULL);int start = 0;int end = 1;int pos = end;while (start != end){Node *node = nvec[start];cout << node->item << " ";start++;if (node->left){nvec.push_back(node->left);snode.push(node->left);end++;}if (node->right){nvec.push_back(node->right);snode.push(node->right);end++;}if (start == pos){pos = end;snode.push(NULL);cout << endl;}}}void putNode2Stack2(Node *root, stack<Node *> &snode){if (root == NULL)return;vector<Node *> nvec;nvec.push_back(root);snode.push(root);snode.push(NULL);int start = 0;int end = 1;int pos = end;while (start != end){Node *node = nvec[start];cout << node->item << " ";start++;if (node->right){nvec.push_back(node->right);snode.push(node->right);end++;}if (node->left){nvec.push_back(node->left);snode.push(node->left);end++;}if (start == pos){pos = end;snode.push(NULL);cout << endl;}}}

利用queue来处理层序遍历

void printLevel(Node *root){if (root == NULL)return;queue<Node *> nqueue;nqueue.push(root);int start = 0;int last = 1;int tempLast = last;while (!nqueue.empty()){Node *top = nqueue.front();nqueue.pop();cout << top->data << " ";if (top->left){nqueue.push(top->left);last++;}if (top->right){nqueue.push(top->right);last++;}start++;if (start == tempLast){tempLast = last;cout << endl;}}}

关于层序遍历，可以出现一个变种题目，那就是NULL指针打印成空格

void printLevel(Node *root){if (root == NULL)return;queue<Node *> qnode;qnode.push(root);int count = 1;int start = 0;int tempCount = 0;while (!qnode.empty()){tempCount = 0;while (start != count){Node *top = qnode.front();if (top != NULL){if (top->left){qnode.push(top->left);tempCount++;}else if (top->left == NULL && top->right != NULL){qnode.push(NULL);tempCount++;}if (top->right){qnode.push(top->right);tempCount++;}else if (top->left != NULL && top->right == NULL){qnode.push(NULL);tempCount++;}}start++;if (top != NULL)cout << top->data;elsecout << " ";qnode.pop(); }count = tempCount;start = 0;cout << endl;}}