题目要求:
从上往下打印出二叉树的每个结点,同一层的结点按照从左往右的顺序打印。
例如输入下图的二叉树,则依次打印出10,5,12,4,7
10 / 5 12 / 4 7
参考资料:剑指offer第23题、编程之美3.10
题目分析:
使用一个辅助队列,队列中的每个元素都是一个指向结点的指针,用递归的方法把二叉树每层依次压入栈中,层与层之间用NULL隔开。
代码实现:

#include <iostream> #include <queue> using namespace std; typedef struct BinaryTree { struct BinaryTree *left,*right; int data; }BinaryTree; void initTree(BinaryTree **p); void PrintTreeByLevel(BinaryTree *root); int main(void) { BinaryTree *root; initTree(&root); cout << "二叉树为:"; PrintTreeByLevel(root); return 0; } //分层遍历二叉树,见编程之美3.10 void PrintTreeByLevel(BinaryTree *root) { if(root==NULL) return; queue<BinaryTree *> Q; Q.push(root); Q.push(0); while(!Q.empty()) { BinaryTree *tmp = Q.front(); Q.pop(); if(tmp) { cout << tmp->data << " "; if(tmp->left) Q.push(tmp->left); if(tmp->right) Q.push(tmp->right); } else if(!Q.empty()) { Q.push(0); } } cout << endl; } // 10 // / // 5 12 // / // 4 7 void initTree(BinaryTree **p) { *p = new BinaryTree; (*p)->data = 10; BinaryTree *tmpNode = new BinaryTree; tmpNode->data = 5; (*p)->left = tmpNode; tmpNode = new BinaryTree; tmpNode->data = 12; (*p)->right = tmpNode; tmpNode->left = NULL; tmpNode->right = NULL; BinaryTree *currentNode = (*p)->left; tmpNode = new BinaryTree; tmpNode->data = 4; currentNode->left = tmpNode; tmpNode->left = NULL; tmpNode->right = NULL; tmpNode = new BinaryTree; tmpNode->data = 7; currentNode->right = tmpNode; tmpNode->left = NULL; tmpNode->right = NULL; }