Java二叉树怎么根据前序和中序推出后续

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根据前序跟中序 => 后序

#include#include#includeusing namespace std;struct BTreeNode{    int _value;    BTreeNode*_left;    BTreeNode*_right;};//解法一BTreeNode* RebuildCode(int * PreStart, int *PreEnd, int *InStart, int *InEnd){    BTreeNode *root = new BTreeNode();    //新建节点root保存前序第一个节点为根结点    root->_value = PreStart[0];    root-> _left = NULL;    root->_right  = NULL;    if(InStart == InEnd && *InStart == *InEnd)    {        return root;    }    //据此节点找到中序遍历此节点的位置    int *rootIn = InStart;    while(*PreStart != *rootIn)    {        rootIn++;    }    //左子树的长度    int leftlen = rootIn-InStart;    //重建左子树    if(leftlen > 0)    {        root->_left = RebuildCode( PreStart+1, PreStart+leftlen, InStart,InStart+leftlen-1);    }    //重建右子树    if(InStart+leftlen < InEnd)    {        root->_right = RebuildCode( PreStart+leftlen+1, PreEnd,InStart+leftlen+1, InEnd);    }    return root;}BTreeNode* RebuildTree(int *PreOrder, int *InOrder, int len){    //首先判断边界条件    if(PreOrder == NULL || InOrder == NULL || len <= 0)    {        return NULL;    }    else    {        return RebuildCode(PreOrder, PreOrder+len-1, InOrder, InOrder+len-1);    }}//后序遍历输出二叉树序列void PostOrder(BTreeNode *root){    if(root->_left != NULL)    {        PostOrder(root->_left);    }     if(root->_right != NULL)    {        PostOrder(root->_right);    }    if(root != NULL)    {        cout<_value<<" ";    }}int main(){    int  PreOrder[8] = {1,2,4,7,3,5,6,8};    int  InOrder[8] = {4,7,2,1,5,3,8,6};    PostOrder(RebuildTree(PreOrder, InOrder, 8));    cout<

根据后序跟中序 =>前序
#include#include#includeusing namespace std;struct BTreeNode{    int _value;    BTreeNode*_left;    BTreeNode*_right;};//解法一BTreeNode* RebuildCode(int * PostStart, int *PostEnd, int *InStart, int *InEnd){    BTreeNode *root = new BTreeNode();    //新建节点root保存前序第一个节点为根结点    root->_value = PostEnd[0];    root-> _left = NULL;    root->_right  = NULL;    if(InStart == InEnd && *InStart == *InEnd)    {        return root;    }    //据此节点找到中序遍历此节点的位置    int *rootIn = InStart;    while(*PostEnd != *rootIn)    {        rootIn++;    }    //左子树的长度    int leftlen = rootIn-InStart;    //重建左子树    if(leftlen > 0)    {        root->_left = RebuildCode( PostStart, PostStart+leftlen-1, InStart,InStart+leftlen-1);    }    //重建右子树    if(InStart+leftlen < InEnd)    {        root->_right = RebuildCode( PostStart+leftlen, PostEnd-1, InStart+leftlen+1, InEnd);    }    return root;}BTreeNode* RebuildTree(int *PostOrder, int *InOrder, int len){    //首先判断边界条件    if(PostOrder == NULL || InOrder == NULL || len <= 0)    {        return NULL;    }    else    {        return RebuildCode(PostOrder, PostOrder+len-1, InOrder, InOrder+len-1);    }}//后序遍历输出二叉树序列void PreOrder(BTreeNode *root){    if(root != NULL)    {        cout<_value<<" ";    }    if(root->_left != NULL)    {        PreOrder(root->_left);    }     if(root->_right != NULL)    {        PreOrder(root->_right);    }}int main(){    int  PostOrder[8] = {7,4,2,5,8,6,3,1};    int  InOrder[8] = {4,7,2,1,5,3,8,6};    PreOrder(RebuildTree(PostOrder, InOrder, 8));    cout<
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