1 問題
按層打印樹節點,比如我們有樹如下
2
3 5
1 4 2 3
這樣打印:2 3 5 1 4 2 3
2 分析
隊列:先進後出,這裏我們先打印2,然後再打印3和5,我們這裏可以使用隊列,我們先把2入隊列,然後我們需要彈出這2節點,先打印隊列最前面的節點,然後再把這個節點的的左右節點都入隊列,然後再彈出最前面的節點,也就是3了,打印出來了就要彈出這個節點,我們希望下次彈出來最前面的節點纔是我們需要打印的,然後再一次把這個彈出的節點左右節點分別入隊列,以次類推,然後循環的條件是隊列爲空。
3 代碼實現
#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);
while(queue.size())
{
Node *node = queue.front();
std::cout << node->value << std::endl;
queue.pop();
if (node->left)
queue.push(node->left);
if (node->right)
queue.push(node->right);
}
}
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;
}
4 運行結果
2
3
5
1
4
2
3