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
代码：
void Level_order ( Tree T, void (*visit)(Tree ThisNode) ) { Tree s[1000]; int head = 0,tail = 0; if(T)s[tail ++] = T; while(head < tail) { if(s[head] -> Left)s[tail ++] = s[head] -> Left; if(s[head] -> Right)s[tail ++] = s[head] -> Right; (*visit)(s[head ++]); } }