从指定节点处拆分二叉树

如T，直接上码子：程序中用到了建树函数（不会搞模板，每次都得重新写，5……%>_<%），前序遍历以验证是否正确拆分开，原树用来保存拆分后剩下的树，newtree则是拆分出来的新树。。。

#include<iostream>#include<stack>using namespace std;typedef struct tr{int data;struct tr *left,*right; }tre,*tree;stack<tree>st;tree creat(tree root,int val){tree newtree;if(NULL==root){newtree=new tre;newtree->data=val;newtree->left=newtree->right=NULL;return newtree;}if(val<=root->data)root->left=creat(root->left,val);elseroot->right=creat(root->right,val);return root;}tree split(tree &root,int val)//指定从某节点处拆开原二叉树{tree newtree;if(NULL!=root){if(root->data==val){newtree=root;root=NULL;return newtree;}else{newtree=split(root->left,val);//拆分左子树if(!newtree)newtree=split(root->right,val);//拆分右子树return newtree;}}elsereturn NULL;}void lbr(tree root){while(!st.empty() || NULL!=root){if(NULL!=root)//左子树入栈{st.push(root);root=root->left;}else{root=st.top();st.pop();printf("%d ",root->data);root=root->right;}}}int main(){int n;tree tmp,newtree;tmp=NULL;while(scanf("%d",&n),n+1)tmp=creat(tmp,n);scanf("%d",&n);newtree=split(tmp,n);lbr(tmp);printf("/n");lbr(newtree);return false;}