面试题32（一）：不分行从上往下打印二叉树（层次遍历二叉树）

一、题目

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到右的顺序打印。

二、关键

1.使用队列deque。

三、解释

1.解题思想：从上到下打印二叉树的规律：每次打印一个节点的时候，如果该节点有子节点，则把该节点的子节点放到一个队列的末尾。接下来到队列的头部取出最早进入队列的节点，重复上面的打印操作，直至队列中所有的节点都被打印出来。

四、代码

#include <cstdio>
#include "..\Utilities\BinaryTree.h"
#include <deque>

void PrintFromTopToBottom(BinaryTreeNode* pRoot)
{
    if(pRoot == nullptr)
        return;

    std::deque<BinaryTreeNode *> dequeTreeNode;

    dequeTreeNode.push_back(pRoot);

    while(dequeTreeNode.size())
    {
        BinaryTreeNode *pNode = dequeTreeNode.front();
        dequeTreeNode.pop_front();

        printf("%d ", pNode->m_nValue);

        if(pNode->m_pLeft)
            dequeTreeNode.push_back(pNode->m_pLeft);

        if(pNode->m_pRight)
            dequeTreeNode.push_back(pNode->m_pRight);
    }
}

// ====================测试代码====================
void Test(char* testName, BinaryTreeNode* pRoot)
{
    if(testName != nullptr)
        printf("%s begins: \n", testName);

    PrintTree(pRoot);

    printf("The nodes from top to bottom, from left to right are: \n");
    PrintFromTopToBottom(pRoot);

    printf("\n\n");
}

//            10
//         /      \
//        6        14
//       /\        /\
//      4  8     12  16
void Test1()
{
    BinaryTreeNode* pNode10 = CreateBinaryTreeNode(10);
    BinaryTreeNode* pNode6 = CreateBinaryTreeNode(6);
    BinaryTreeNode* pNode14 = CreateBinaryTreeNode(14);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode8 = CreateBinaryTreeNode(8);
    BinaryTreeNode* pNode12 = CreateBinaryTreeNode(12);
    BinaryTreeNode* pNode16 = CreateBinaryTreeNode(16);

    ConnectTreeNodes(pNode10, pNode6, pNode14);
    ConnectTreeNodes(pNode6, pNode4, pNode8);
    ConnectTreeNodes(pNode14, pNode12, pNode16);

    Test("Test1", pNode10);

    DestroyTree(pNode10);
}

//               5
//              /
//             4
//            /
//           3
//          /
//         2
//        /
//       1
void Test2()
{
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);

    ConnectTreeNodes(pNode5, pNode4, nullptr);
    ConnectTreeNodes(pNode4, pNode3, nullptr);
    ConnectTreeNodes(pNode3, pNode2, nullptr);
    ConnectTreeNodes(pNode2, pNode1, nullptr);

    Test("Test2", pNode5);

    DestroyTree(pNode5);
}

// 1
//  \
//   2
//    \
//     3
//      \
//       4
//        \
//         5
void Test3()
{
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
    BinaryTreeNode* pNode2 = CreateBinaryTreeNode(2);
    BinaryTreeNode* pNode3 = CreateBinaryTreeNode(3);
    BinaryTreeNode* pNode4 = CreateBinaryTreeNode(4);
    BinaryTreeNode* pNode5 = CreateBinaryTreeNode(5);

    ConnectTreeNodes(pNode1, nullptr, pNode2);
    ConnectTreeNodes(pNode2, nullptr, pNode3);
    ConnectTreeNodes(pNode3, nullptr, pNode4);
    ConnectTreeNodes(pNode4, nullptr, pNode5);

    Test("Test3", pNode1);

    DestroyTree(pNode1);
}

// 树中只有1个结点
void Test4()
{
    BinaryTreeNode* pNode1 = CreateBinaryTreeNode(1);
    Test("Test4", pNode1);

    DestroyTree(pNode1);
}

// 树中没有结点
void Test5()
{
    Test("Test5", nullptr);
}

int main(int argc, char* argv[])
{
    Test1();
    Test2();
    Test3();
    Test4();
    Test5();

   return 0;
}