[转]根据二叉树的先序、中序遍历结果重建二叉树

 

    #include <stdlib.h> 
    #include <stdio.h> 
     
    typedef struct TNode 
    { 
        int value; 
        TNode* lchild; 
        TNode* rchild; 
    }TNode,*BTree;  
     
    //根据先序遍历、中序遍历构建二叉树 
    BTree rebuild(int preOrder[],int startPre,int endPre,int inOrder[],int startIn,int endIn) 
    { 
        //先序遍历和中序遍历长度应相等 
        if (endPre - startPre != endIn - startIn) return NULL; 
        //起始位置不应大于末尾位置 
        if (startPre > endPre) return NULL; 
        //先序遍历的第一个元素为根节点 
        BTree tree = (BTree)malloc(sizeof(TNode)); 
        tree->value = preOrder[startPre]; 
        tree->lchild = NULL; 
        tree->rchild = NULL; 
        //先序遍历和中序遍历只有一个元素时，返回该节点 
        if (startPre == endPre) return tree; 
        //在中序遍历中找到根节点 
        int index,length; 
        for (index=startIn;index<=endIn;index++) 
        { 
            if (inOrder[index] == preOrder[startPre]) break; 
        } 
        //若未找到，返回空 
        if (index > endIn) return NULL; 
        //有左子树，递归调用构建左子树 
        if (index > startIn)  
        { 
            length = index-startIn; 
            tree->lchild = rebuild(preOrder,startPre+1,startPre+1+length-1,inOrder,startIn,startIn+length-1); 
        } 
        //有右子树，递归调用构建右子树 
        if (index < endIn)  
        { 
            length = endIn - index; 
            tree->rchild = rebuild(preOrder,endPre-length+1,endPre,inOrder,endIn-length+1,endIn); 
        } 
        return tree; 
    } 
     
    //后序遍历二叉树 
    void postTraverse(BTree tree) 
    { 
        if (tree->lchild != NULL) postTraverse(tree->lchild); 
        if (tree->rchild != NULL) postTraverse(tree->rchild); 
        printf("%d ",tree->value); 
    } 
     
    int main() 
    { 
        int preOrder[] = {1,2,4,5,3,6}; 
        int inOrder[] = {4,2,5,1,6,3}; 
        BTree tree = rebuild(preOrder,0,5,inOrder,0,5); 
        postTraverse(tree); 
        printf("\n"); 
        return 0; 
    }