二叉树的层次遍历

8

/ \

6   10

/ \  / \

5  7 9   11

层次遍历输出结果： 8,6,10,5,7,9,11

#include <deque>

using std::deque;

#include <iostream>

using namespace std;

template<typename T>

struct TreeNode

{

public:

TreeNode(T value);

void setLeftChild( TreeNode *lChild );

void setRightChild( TreeNode *rChild );

TreeNode *getLeftChild ();

TreeNode *getRightChild();

void print();

private:

TreeNode *mLchild;

TreeNode *mRchild;

T mValue;

};

template<typename T>

TreeNode<T>::TreeNode(T value):mValue(value),mLchild(0),mRchild(0)

{

}

template<typename T>

void TreeNode<T>::setLeftChild( TreeNode *lChild )

{

mLchild = lChild;

}

template<typename T>

void TreeNode<T>::setRightChild( TreeNode *rChild )

{

mRchild = rChild;

}

template<typename T>

TreeNode<T> *

TreeNode<T>::getLeftChild ( )

{

return mLchild;

}

template<typename T>

TreeNode<T> *

TreeNode<T>::getRightChild ( )

{

return mRchild;

}

template<typename T>

void

TreeNode<T>::print()

{

cout << mValue <<endl;

}

template<typename T>

TreeNode<T> *

BuildTree()

{

TreeNode<int> *r8 = new TreeNode<int> ( 8 );

TreeNode<int> *r6 = new TreeNode<int> ( 6 );

TreeNode<int> *r10 = new TreeNode<int> ( 10 );

TreeNode<int> *r5 = new TreeNode<int> ( 5 );

TreeNode<int> *r7 = new TreeNode<int> ( 7 );

TreeNode<int> *r9 = new TreeNode<int> ( 9 );

TreeNode<int> *r11 = new TreeNode<int> ( 11 );

r8->setLeftChild(r6);

r8->setRightChild(r10);

r6->setLeftChild(r5);

r6->setRightChild(r7);

r10->setLeftChild(r9);

r10->setRightChild(r11);

return r8;

}

template<typename T>

void printTree(TreeNode<T> *node)

{

deque<TreeNode<T>*> fifo;

fifo.push_back(node);

while ( fifo.size() )

{

TreeNode<T> *front = fifo.front();

fifo.pop_front();

front->print();

if ( front->getLeftChild() )

    {

fifo.push_back(front->getLeftChild());

}

if ( front->getRightChild() )

    {

fifo.push_back(front->getRightChild());

}

}

}

int main()

{

TreeNode<int> *node = BuildTree<int>();

printTree(node);

return 0;

}