层次遍历二叉树（编程之美3.10）

问题（假定根节点位于第0层）

1. 层次遍历二叉树（每层换行分开）

2. 层次遍历二叉树指定的某层

例如

上图中

1.

1
2 3
4 5 6
7 8

2.

第三层
7 8

可以看出得出第二问的解，第一问迎刃而解了，所以从问题二下手

 

分析与解

1. 层次遍历二叉树指定的某层

可以得出这样的一个结论：遍历二叉树的第k层，相当于遍历二叉树根节点的左右子树的第k-1层。这样一直遍历下去，直到k=0时，输出节点即可。

参考代码

int PrintNodeAtLevel(BiTree root, int level)
{
    if(!root || level < 0)
        return 0;
    else if(level == 0)
    {
        cout << root->data << endl;
        return 1;
    }
    else
        return PrintNodeAtLevel(root->left, level - 1) + PrintNodeAtLevel(root->right, level - 1);
}

完整执行代码

#include<iostream>
using namespace std;
typedef struct BiTNode
{
    int data;
    BiTNode *left;
    BiTNode *right;
}BiTNode, *BiTree;

void createTree(BiTree &root)
{
    BiTree left1 = new(BiTNode);
    BiTree right1 = new(BiTNode);
    
    left1->data = 1;
    left1->left = NULL;
    left1->right = NULL;
    right1->data = 2;
    right1->left = NULL;
    right1->right = NULL;

    root->left = left1;
    root->right = right1;


    BiTree left2 = new(BiTNode);
    left2->data = 3;
    left2->left = NULL;
    left2->right = NULL;
    BiTree right2 = new(BiTNode);
    right2->data = 4;
    right2->left = NULL;
    right2->right = NULL;
    left1->left = left2;
    left1->right = right2;

    BiTree left3 = new(BiTNode);
    left3->data = 5;
    left3->left = NULL;
    left3->right = NULL;
    BiTree right3 = new(BiTNode);
    right3->data = 6;
    right3->left = NULL;
    right3->right = NULL;
    left2->left = left3;
    left2->right = right3;
}

void deleteTree(BiTree root)
{
    if(root)
    {
        deleteTree(root->left);
        deleteTree(root->right);
        delete(root);
        root = NULL;
    }
}

int PrintNodeAtLevel(BiTree root, int level)
{
    if(!root || level < 0)
        return 0;
    else if(level == 0)
    {
        cout << root->data << endl;
        return 1;
    }
    else
        return PrintNodeAtLevel(root->left, level - 1) + PrintNodeAtLevel(root->right, level - 1);
}

int main()
{
    BiTree root = new(BiTNode);
    root->data = 0;
    root->right = root->left = NULL;
    createTree(root);    
    cout << "Level 0:" << endl;
    PrintNodeAtLevel(root, 0);
    cout << "-------------------" << endl;
    cout << "Level 1:" << endl;
    PrintNodeAtLevel(root, 1);
    cout << "-------------------" << endl;
    cout << "Level 2:" << endl;
    PrintNodeAtLevel(root, 2);
    cout << "-------------------" << endl;
    cout << "Level 3:" << endl;
    PrintNodeAtLevel(root, 3);
    cout << "-------------------" << endl;
    deleteTree(root);
}

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