笔试题:二叉树按层遍历&添加兄弟指针&求LCA&排序二叉树的查找

1.二叉树按层遍历

2.二叉树添加兄弟指针

3.在二叉树中查找LCA（最近公共祖先） 

3.在排序二叉树中找到大于N且最接近N的数 

// PrintByLevel.cpp : Defines the entry point for the console application.
// Author : yangyh

#include "stdafx.h"
#include <iostream>
#include <queue>
using namespace std;
typedef struct _node_st{
	int value;
	_node_st* pLeft;
	_node_st* pRight;
	_node_st* pNext;
	_node_st(){
		pLeft=NULL;
		pRight=NULL;
		pNext=NULL;
	}
}Node,*NodePtr;

void visitNode(int level,NodePtr node){
	printf("Level% d: %d\n",level,node->value);
}

//在二叉查找树中找到比n大的最接近n的数
int findNum(NodePtr root,int n){

	if(!root)return -1;
	if(n<root->value)
	{
		//如果左子树为空，则root->value即为返回值
		if(!root->pLeft)return root->value;
		//否则在左子树返回结果与root->value对比，取较小值
		int left = findNum(root->pLeft,n);
		if(left!=-1){
			return left<root->value?(left):(root->value);
		}
		return root->value;
	
	}
	return findNum(root->pRight,n);
}

//寻找a b的最近公共祖先，LCA满足一个特征：a,b分别在它的左右子树
//求LCA的解法2：分别用栈保存遍历到a,b的路径，则问题转换为求两链表的交点，
//假设栈A长度为LA，栈B LB，如果LA>LB,则A指针先走LA-LB步，之后同时走，A=B时即为交点
NodePtr LCA(NodePtr root,NodePtr a,NodePtr b){

	if(root==NULL)return NULL;
	if(a == b)
		return a;
	if(a==root||b==root)return root;

	NodePtr pleft = LCA(root->pLeft,a,b);  //在左子树找到a或b
	NodePtr pright = LCA(root->pRight,a,b);//在右子树找到a或b
	//a,b分别分布在左右子树，则此根结点即LCA
	if(pleft&&pright)
		return root;
	//右子树中找不到a,b，则在左子树中找到的LCA即是root下的LCA
	if(pleft!=NULL)
		return pleft;
	//左子树中找不到a,b，则在右子树中找到的LCA即是root下的LCA
	if(pright!=NULL)
		return pright;

	return NULL;

}
//给二叉树中的同层节点添加兄弟指针
int addSiblingPtr(NodePtr root,void (*visit)(int,NodePtr)){
	if(root==NULL)
		return 0;
	queue<NodePtr> q;
	q.push(root);
	int level = 1;
	while(!q.empty()){
		int count = q.size();
		NodePtr front=NULL;
		NodePtr next=NULL,node;
		while(count){
			node = q.front();
			if(node->pLeft)q.push(node->pLeft);
			if(node->pRight)q.push(node->pRight);
			q.pop();
			visit(level,node);
			
			if(front==NULL)
				front = node;
			else{
				front->pNext=node;
				front = node;
			}
			count--;
		}
		
		level++;
		node->pNext=NULL;
	}
	return 1;
}

int main(int argc, char* argv[])
{
	//初始二叉树
	Node nodes[10];
	for(int i=0;i<sizeof(nodes)/sizeof(Node);i++){
		
		nodes[i].value = i;
		
	}
	//树结构  1
	//       / \
	//      2   3
	//     / \ / \
	//     4 5 6 7
	nodes[1].pLeft=&nodes[2];
	nodes[1].pRight=&nodes[3];
	nodes[2].pLeft=&nodes[4];
	nodes[3].pLeft=&nodes[6];
	nodes[2].pRight=&nodes[5];
	nodes[3].pRight=&nodes[7];

	addSiblingPtr(&nodes[1],visitNode);

	printf("%d\n",nodes[2].pNext->value);
	printf("%d\n",nodes[4].pNext->value);
	printf("%d\n",nodes[5].pNext->value);
	printf("%d\n",nodes[6].pNext->value);

	printf("%d\n",LCA(&nodes[1],&nodes[7],&nodes[4])->value);
    printf("%d\n",LCA(&nodes[1],&nodes[6],&nodes[4])->value);
	printf("%d\n",LCA(&nodes[1],&nodes[7],&nodes[6])->value);

	//树结构  8
	//       / \
	//      5   13
	//     / \  / \
	//    4  7 11 17
	nodes[1].value = 8;
	nodes[2].value = 5;nodes[3].value =13;
	nodes[4].value = 4;nodes[5].value = 7;
	nodes[6].value = 11;nodes[7].value = 17;

	printf("%d\n",findNum(&nodes[1],12));
	printf("%d\n",findNum(&nodes[1],-2));
	printf("%d\n",findNum(&nodes[1],6));
	printf("%d\n",findNum(&nodes[1],8));
	printf("%d\n",findNum(&nodes[1],10));
	return 0;
}