《剑指offer》第三十二题I：不分行从上往下打印二叉树

Posted 2021-03-03 zsy-blog

// 面试题32（一）：不分行从上往下打印二叉树
// 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。

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

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

    std::deque<BinaryTreeNode *> dequeTreeNode; //STL队列

    dequeTreeNode.push_back(pRoot);

    while (dequeTreeNode.size())
    {
        //弹出队列头元素并打印
        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(const char* testName, BinaryTreeNode* pRoot)
{
    if (testName != nullptr)
        printf("%s begins: 
", testName);

    PrintTree(pRoot);

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

    printf("

");
}

//            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;
}

测试代码

分析：队列先进先出。

/*
struct TreeNode {
    int val;
    struct TreeNode *left;
    struct TreeNode *right;
    TreeNode(int x) :
            val(x), left(NULL), right(NULL) {
    }
};*/
class Solution {
public:
    vector<int> PrintFromTopToBottom(TreeNode* root) {
        
        std::deque<TreeNode*> dequeTreeNode;
        std::vector<int> printTreeNode;
        
        if (root == nullptr)
            return printTreeNode;
        
        dequeTreeNode.push_back(root);
        
        while (dequeTreeNode.size())
        {
            TreeNode* pNode = dequeTreeNode.front();
            dequeTreeNode.pop_front();
            
            printTreeNode.push_back(pNode->val);
            
            if (pNode->left)
                dequeTreeNode.push_back(pNode->left);
            if (pNode->right)
                dequeTreeNode.push_back(pNode->right);
        }
        return printTreeNode;
    }
};

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