6-11 Level-order Traversal（25 point(s)）

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本文介绍了一种实现二叉树层序遍历的方法，采用队列来辅助完成节点的逐层访问，并提供了一个具体的C语言实现示例。

6-11 Level-order Traversal（25 point(s)）

Write a routine to list out the nodes of a binary tree in "level-order". List the root, then nodes at depth 1, followed by nodes at depth 2, and so on. You must do this in linear time.

Format of functions:

void Level_order ( Tree T, void (*visit)(Tree ThisNode) );

where void (*visit)(Tree ThisNode) is a function that handles ThisNode being visited by Level_order, and Tree is defined as the following:

typedef struct TreeNode *Tree;
struct TreeNode {
    ElementType Element;
    Tree  Left;
    Tree  Right;
};

Sample program of judge:

#include <stdio.h>
#include <stdlib.h>

#define MaxTree 10 /* maximum number of nodes in a tree */
typedef int ElementType;

typedef struct TreeNode *Tree;
struct TreeNode {
    ElementType Element;
    Tree  Left;
    Tree  Right;
};

Tree BuildTree(); /* details omitted */
void PrintNode( Tree NodePtr )
{
   printf(" %d", NodePtr->Element);
}

void Level_order ( Tree T, void (*visit)(Tree ThisNode) );

int main()
{
    Tree T = BuildTree();
    printf("Level-order:");
    Level_order(T, PrintNode);
    return 0;
}

/* Your function will be put here */

Sample Output (for the tree shown in the figure):

Level-order: 3 5 6 1 8 10 9

code：
void Level_order ( Tree T, void (*visit)(Tree ThisNode) ){
    Tree q[1000];
    Tree p;
    int head=0,tail=0;
    if(!T)return;
    if(T){
        q[tail++] = T;
        while(head!=tail){
            p = q[head++];
            visit(p);
            if(p->Left)q[tail++]=p->Left;
            if(p->Right)q[tail++]=p->Right;
        }
    }
}//同上一题用队列进行层序遍历