/*求二叉樹葉子節(jié)點(diǎn)個數(shù) -- 采用遞歸和非遞歸方法
經(jīng)調(diào)試可運(yùn)行源碼及分析如下：
***/
#include <stdlib.h>
#include <iostream>
#include <stack>
using std::cout;
using std::cin;
using std::endl;
using std::stack;
/*二叉樹結(jié)點(diǎn)定義*/
typedef struct BTreeNode
{
  char elem;
  struct BTreeNode *pleft;
  struct BTreeNode *pright;
}BTreeNode;
/*
求二叉樹葉子節(jié)點(diǎn)數(shù)
葉子節(jié)點(diǎn)：即沒有左右子樹的結(jié)點(diǎn)
遞歸方式步驟：
如果給定節(jié)點(diǎn)proot為NULL，則是空樹，葉子節(jié)點(diǎn)為0，返回0；
如果給定節(jié)點(diǎn)proot左右子樹均為NULL，則是葉子節(jié)點(diǎn)，且葉子節(jié)點(diǎn)數(shù)為1，返回1；
如果給定節(jié)點(diǎn)proot左右子樹不都為NULL，則不是葉子節(jié)點(diǎn)，以proot為根節(jié)點(diǎn)的子樹葉子節(jié)點(diǎn)數(shù)=proot左子樹葉子節(jié)點(diǎn)數(shù)+proot右子樹葉子節(jié)點(diǎn)數(shù)。
/*遞歸實(shí)現(xiàn)求葉子節(jié)點(diǎn)個數(shù)*/
int get_leaf_number(BTreeNode *proot)
{
  if(proot == NULL)
    return 0;
  if(proot->pleft == NULL && proot->pright == NULL)
    return 1;
  return (get_leaf_number(proot->pleft) + get_leaf_number(proot->pright));
}
/*非遞歸：本例采用先序遍歷計算
判斷當(dāng)前訪問的節(jié)點(diǎn)是不是葉子節(jié)點(diǎn)，然后對葉子節(jié)點(diǎn)求和即可。
 **/
int preorder_get_leaf_number(BTreeNode* proot)
{
  if(proot == NULL)
    return 0;
  int num = 0;
  stack <BTreeNode*> st;
  while (proot != NULL || !st.empty())
  {
    while (proot != NULL)
    {
      cout << "節(jié)點(diǎn)：" << proot->elem << endl;
      st.push(proot);
      proot = proot->pleft;
    }
    if (!st.empty())
    {
      proot = st.top();
      st.pop();
      if(proot->pleft == NULL && proot->pright == NULL)
        num++;
      proot = proot -> pright;
    }
  }
  return num;
}
/*初始化二叉樹根節(jié)點(diǎn)*/
BTreeNode* btree_init(BTreeNode* &bt)
{
  bt = NULL;
  return bt;
}
/*先序創(chuàng)建二叉樹*/
void pre_crt_tree(BTreeNode* &bt)
{
  char ch;
  cin >> ch;
  if (ch == '#')
  {
    bt = NULL;
  }
  else
  {
    bt = new BTreeNode;
    bt->elem = ch;
    pre_crt_tree(bt->pleft);
    pre_crt_tree(bt->pright);
  }
}
int main()
{
  int tree_leaf_number = 0;
  BTreeNode *bt;
  btree_init(bt);//初始化根節(jié)點(diǎn)
  pre_crt_tree(bt);//創(chuàng)建二叉樹
  tree_leaf_number = get_leaf_number(bt);//遞歸
  cout << "二叉樹葉子節(jié)點(diǎn)個數(shù)為:" << tree_leaf_number << endl;
  cout << "非遞歸先序遍歷過程如下：" << endl;
  tree_leaf_number = preorder_get_leaf_number(bt);//非遞歸
  cout << "二叉樹葉子節(jié)點(diǎn)個數(shù)為:" << tree_leaf_number << endl;
  system("pause");
  return 0;
}
/*

運(yùn)行結(jié)果：
a b c # # # d e # # f # #
---以上為輸入---
---以下為輸出---
二叉樹葉子節(jié)點(diǎn)個數(shù)為:3
非遞歸遍歷過程如下：
節(jié)點(diǎn)：a
節(jié)點(diǎn)：b
節(jié)點(diǎn)：c
節(jié)點(diǎn)：d
節(jié)點(diǎn)：e
節(jié)點(diǎn)：f
二叉樹葉子節(jié)點(diǎn)個數(shù)為:3
請按任意鍵繼續(xù). . .

本例創(chuàng)建的二叉樹形狀：
    a
  b    d  
c     e  f
*/