【数据结构·考研】二叉树的结点个数

二叉树的结点个数

非递归的求法很简单，每出队一个结点，count++。递归的解法，从上自下遍历到最底端，返回左右子树结点数+1，代表左右子树的结点再加上父结点的个数。

代码如下：

#include<iostream>
#include<queue>
#include<vector>
using namespace std;

typedef struct node{
	char val;
	struct node* left;
	struct node* right;
}TreeNode,*Tree;

//递归 
/*递归到树的最底部，返回左右子树的结点数再+1，表示左右子树再加上父结点*/ 
int Count(Tree& t){
    if(t == NULL)  return 0; 
    return Count(t->left) + Count(t->right) + 1;
}

//非递归 
int levelOrder(Tree& t) {
    if(t == NULL) return 0;
    int count = 0; //结点总数 
    queue<TreeNode*> q;
    q.push(t);
    while(!q.empty()){ 
        TreeNode* s = q.front();
		count ++; 
        q.pop();
        if(s->left) q.push(s->left);
        if(s->right) q.push(s->right);
    } 
    return count;
}

void CreateTree(Tree& t){
	char x;
	cin>>x;
	if(x == '#') t = NULL; 
	else{
		t = new TreeNode; 
		t->val = x;  
		CreateTree(t->left); 
		CreateTree(t->right); 
	}
} 

int main(){
	Tree t;
	CreateTree(t);
	/*
	   a b d # # e # # c f # # #
	*/
	cout<<endl<<"非递归:"<<endl;
	cout<<levelOrder(t); 
	cout<<endl<<"递归:"<<endl;
	cout<<Count(t);
}

运行结果：