二叉树的插入,递归遍历和深度节点数

 

二叉树的插入,前序,中序,后序遍历,递归求深度和节点数

 

 

 

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

struct tree_el {
   int val;
   struct tree_el * right, * left;
};

typedef struct tree_el node;

void insert(node ** tree, node * item) {
   if(!(*tree))
    {
      *tree = item;
      return;
    }
   if(item->val<=(*tree)->val)
      insert(&(*tree)->left, item);
   else if(item->val>(*tree)->val)
      insert(&(*tree)->right, item);
}

void InOrd(node * tree)
{
   if(tree->left)
        InOrd(tree->left);
   printf("%d ",tree->val);
   if(tree->right)
        InOrd(tree->right);
}

void PreOrd(node *tree)
{
    printf("%d ",tree->val);
    if(tree->left)
        PreOrd(tree->left);
   if(tree->right)
        PreOrd(tree->right);
}

void PostOrd(node *tree)
{
    if(tree->left)
        PostOrd(tree->left);
   if(tree->right)
        PostOrd(tree->right);
    printf("%d ",tree->val);
}

int TreeDepth(node* tree)
{
      // the depth of a empty tree is 0
      if(!tree)
            return 0;

      // the depth of left sub-tree
      int nLeft = TreeDepth(tree->left);
      // the depth of right sub-tree
      int nRight = TreeDepth(tree->right);

      // depth is the binary tree
      return (nLeft > nRight) ? (nLeft + 1) : (nRight + 1);
}

int BTreeNodeCount(node *tree)
{
    if(tree == 0)
        return 0;
    else
    {
        int l = BTreeNodeCount(tree->left);
        int r = BTreeNodeCount(tree->right);
        return l + r + 1;
    }
}

int main() {
   node * curr, * root;
   int i;
   int data;
   int lenth;

   root = NULL;
   printf("请输入要插入的元素,按Ctrl + Z结束输入\n");
   while(scanf("%d",&data)!= EOF)
   {
       curr = (node *)malloc(sizeof(node));
      curr->left = curr->right = NULL;
      curr->val = data;
      insert(&root, curr);
   }
   printf("先序遍历 : ");
   PreOrd(root);

   printf("\n中序遍历 : ");
   InOrd(root);

   printf("\n后序遍历 : ");
    PostOrd(root);

    printf("\n树的深度为 : %d\n", TreeDepth(root));

    printf("树的节点数为:%d\n", BTreeNodeCount(root));

    return 0;
}
posted on 2012-04-23 00:50  genslow  阅读(317)  评论(0编辑  收藏  举报