面试题6

1 题目描述

根据先序序列和中序序列构建二叉树

2 算法描述

1 遍历先序序列，获得根节点 R
2 在中序序列中检索 R，确定 R 左子树与右子树的范围
3 递归上述过程

C 语言实现

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#include<stdio.h>typedef int ElemType;typedef struct node{    ElemType data;    struct node *lchild,*rchild;}TNode,*Tree;Tree constructCore(int* sp,int* ep,int* si,int* ei){    //先序遍历的第一个结点为根结点    TNode* node = (TNode*)malloc(sizeof(TNode));    node->data=sp[0];    node->lchild=NULL;    node->rchild=NULL;    //遍历 中序序列确定根结点的位置，并确定左子树和右子树范围，构建    int index=0;    int flag=0;    while(index<=ei-si){        if(si[index++]==sp[0]){            flag=1;            index--;            break;        }    }    if(flag==0) {        printf("前序序列与中序序列无法形成二叉树\n");        return NULL;    }    //index 值为根结点在中序序列中的索引    //大于 0 说明存在左子树    if(index>0){        //递归构建左子树        node->lchild=constructCore(sp+1,sp+index,si,si+index-1);    }    //存在右子树    if(index<ei-si){        node->rchild=constructCore(sp+index+1,ep,si+index+1,ei);    }    return node;}Tree constructTree(int preorder[],int inorder[],int length){    return constructCore(preorder,preorder+length-1,inorder,inorder+length-1);}//前序遍历二叉树void preOrder(TNode* node){    if(node==NULL) return;    else{        printf("%d",node->data);        preOrder(node->lchild);        preOrder(node->rchild);    }}void main(){    int preorder[]={1,2,4,7,3,5,6,8};    int inorder[]={4,7,2,1,5,3,8,6};    Tree tree=NULL;    tree=constructTree(preorder,inorder,8);    preOrder(tree);}

Java 语言实现

package _6;public class ConstructTree {    private static class TreeNode{        int data;        TreeNode lchild;        TreeNode rchild;    }    public static void main(String[] args) {        int[] preorder={1,2,4,7,3,5,6,8};        int[] inorder={4,7,2,1,5,3,8,6};        TreeNode tree=constructTree(preorder,inorder,8);        preOrderTree(tree);    }    public static TreeNode constructTree(int[] preorder,int[] inorder,int length){        if(preorder==null||inorder==null||length<=0){            throw new RuntimeException();        }        return constructCore(preorder,0,length-1,inorder,0,length-1);    }    public static TreeNode constructCore(int[] preorder,  int prestart,int preend,            int inorder[],int instart,int inend){        TreeNode node=new TreeNode();        node.data=preorder[prestart];        node.lchild=node.rchild=null;        int index=0;        boolean flag=false;        while(index<=inend-instart){            if(inorder[instart+index]==node.data){                flag=true;                break;            }            index++;        }        if(!flag){            throw new RuntimeException();        }        //index为 根结点 中序序列中的索引        //判断是否存在左子树        if(index>0){            node.lchild=constructCore(preorder,prestart+1,prestart+index,inorder,instart,instart+index-1);        }        if(index<inend-instart){            node.rchild=constructCore(preorder,prestart+index+1,preend,inorder,instart+index+1,inend);        }        return node;    }    public static void preOrderTree(TreeNode node){        if(node==null) return;        System.out.println(node.data);        preOrderTree(node.lchild);        preOrderTree(node.rchild);    }}