剑指offer之分行从上到下打印二叉树

1 题目

分行从上到下打印二叉树

                  23    5           1  4  2  3

我们打印如下

23     51     4     2     3

2 分析

之前这篇博客写了通过队列按层打印剑指offer之按层打印树节点

现在无非就是还要按照条件打印，第一次打印1个，然后第二次打印2个，第三次打印4个，我们可以这样，搞2个变量，第一个变量表示这行还没有打印的个数，另外一个变量是下列需要打印的个数，如果这一行还没有打印的个数是0了，我们就把下列需要打印值个数赋值给它，然后下一列变量的个数变量赋值为0.

3 代码实现

#include <iostream>
#include <queue>using namespace std;typedef struct Node
{int value;struct Node* left;struct Node* right;
} Node;void layer_print(Node *head)
{if (head == NULL){std::cout << "head is NULL" << std::endl;return;}std::queue<Node *> queue;queue.push(head);//下一列需要打印节点的个数int next_print_count = 0;//当前这一列还需要打印节点的个数int has_print_count = 1;while(queue.size()){Node *node = queue.front();std::cout << node->value << "\t";if (node->left){queue.push(node->left);next_print_count++;}if (node->right) {queue.push(node->right);next_print_count++;}queue.pop();has_print_count--;if (has_print_count == 0){has_print_count = next_print_count;next_print_count = 0;std::cout << std::endl;}} 
}int main()
{/*              2*           3    5           *         1  4  2  3       *       */Node head1, node1, node2, node3, node4, node5, node6;Node head2, node7, node8;head1.value = 2;node1.value = 3;node2.value = 5;node3.value = 1;node4.value = 4;node5.value = 2;node6.value = 3;head1.left = &node1;head1.right = &node2;node1.left = &node3;node1.right = &node4;node2.left = &node5;node2.right = &node6;node3.left = NULL;node3.right = NULL;node4.left = NULL;node4.right = NULL;node5.left = NULL;node5.right = NULL;node6.left = NULL;node6.right = NULL;layer_print(&head1);return 0;
}