queue 的基本操作舉例如下：

queue入隊，如例：q.push(x); 將x 接到隊列的末端。 queue出隊，如例：q.pop(); 彈出隊列的第一個元素，注意，并不會返回被彈出元素的值。 訪問queue隊首元素，如例：q.front()，即最早被壓入隊列的元素。 訪問queue隊尾元素，如例：q.back()，即最后被壓入隊列的元素。 判斷queue隊列空，如例：q.empty()，當隊列空時，返回true。 訪問隊列中的元素個數，如例：q.size() #include<iostream> #include<stack> #include<queue>using namespace std;typedef struct TreeNode {int data;struct TreeNode* lchild;struct TreeNode* rchild; }TreeNode;void pretravel(TreeNode* T) {if (T != NULL){cout << T->data << " ";pretravel(T->lchild);pretravel(T->rchild);} }void intravel(TreeNode* T) {if (T != NULL){intravel(T->lchild);cout << T->data<<" ";intravel(T->rchild);} }void posttravel(TreeNode* T) {if (T != NULL){posttravel(T->lchild);posttravel(T->rchild);cout << T->data << " ";} }/****************************************************/void preOrdtravel(TreeNode* T) //先序非遞歸 {if (!T) return;stack<TreeNode*> s;s.push(T);TreeNode* t;while (!s.empty()){t = s.top(); //出棧并打印cout << t->data << " ";s.pop();if (t->rchild)s.push(t->rchild);if (t->lchild)s.push(t->lchild);}}void inOrdTravel(TreeNode* T) //中序非遞歸 {if (!T)return;TreeNode* cur = T;stack<TreeNode*> S;while (!S.empty() || cur != NULL){if (cur != NULL){S.push(cur);cur = cur->lchild;}else{//出棧并打印cout << S.top()->data << " ";cur = S.top()->rchild;S.pop();}} }void postOrdtravel(TreeNode *T) //后序非遞歸 {if (T == NULL) return;TreeNode *p = T;stack<TreeNode *> S;TreeNode *last = T; //last=NULL（錯誤），必須是T（正確）S.push(p);while (!S.empty()){p = S.top(); //獲取棧頂if ((p->lchild == NULL && p->rchild == NULL) || (p->rchild == NULL && last == p->lchild) || (last == p->rchild)){//打印并出棧，更新lastcout << p->data << " ";last = p;S.pop();}else{if (p->rchild)S.push(p->rchild);if (p->lchild)S.push(p->lchild);}} }void levelTraver(TreeNode* T) //層次遍歷 {if (!T)return;queue<TreeNode*> Q;TreeNode* cur = T;Q.push(cur);while (!Q.empty()){cout << Q.front()->data << " ";cur = Q.front();Q.pop();if (cur->lchild)Q.push(cur->lchild);if (cur->rchild)Q.push(cur->rchild);} }int main() {TreeNode* s1 = new TreeNode;TreeNode* s2 = new TreeNode;TreeNode* s3 = new TreeNode;TreeNode* s4 = new TreeNode;TreeNode* s5 = new TreeNode;TreeNode* s6 = new TreeNode;TreeNode* s7 = new TreeNode;s1->data = 1;s1->lchild = s2;s1->rchild = s5;s2->data = 2;s2->lchild = NULL;s2->rchild = s3;s3->data = 3;s3->lchild = s4;s3->rchild = NULL;s4->data = 4;s4->lchild = NULL;s4->rchild = NULL;s5->data = 5;s5->lchild = s6;s5->rchild = NULL;s6->data = 6;s6->lchild = NULL;s6->rchild = s7;s7->data = 7;s7->lchild = NULL;s7->rchild = NULL;//遍歷cout << "先序遍歷"<<endl;pretravel(s1); cout << endl;preOrdtravel(s1); cout << endl;cout << "中序遍歷" << endl;intravel(s1); cout << endl;inOrdTravel(s1); cout << endl;cout << "后序遍歷" << endl;posttravel(s1); cout << endl;postOrdtravel(s1); cout << endl;cout << "層序遍歷" << endl;levelTraver(s1);return 0;} 《新程序員》：云原生和全面數字化實踐50位技術專家共同創作，文字、視頻、音頻交互閱讀

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