3.9重建二叉树

扩展问题2

如何判定给定的前序遍历和中序遍历的结果是合理的？

 

来源于

http://blog.csdn.net/bvbook/archive/2008/07/25/2710382.aspx

 

 

递归算法实现，分别遍历左右子树，递归中迭代查找左右子树的长度，类似于书中的方法。

#include <iostream>#include <string>using namespace std;bool valid = true;inline int find(char needle, const char* haystack, int start, int end ){ const char* p = haystack + start; int i = 0; int offset = end - start ; while ( p && i < offset ) { if ( *p == needle ) { if ( start != 0 ) return i + start; else return i; } p++, i++; } return -1;}void isValid( const char* preOrder, const char* inOrder, int start, int end ){ if ( !valid ) return ; int position = find( *preOrder, inOrder, start,end ); if ( position == -1 ) { valid = false; return ; } if ( start < position ) { isValid( preOrder + 1, inOrder,start,position ); //在左子树中遍历 } if ( position + 1 < end ) { isValid( preOrder + position + 1, inOrder, position + 1, end ); //在右子树中遍历 }}// Two Simple Test Casesint main(){ string pre1 = "abdefc"; string mid1 = "dbfeac"; string pre2 = "abdefc"; string mid2 = "dcfeab"; isValid(pre1.c_str(),mid1.c_str(),0,mid1.length()); cout << valid << endl; // 输出 true valid = true; isValid(pre2.c_str(),mid2.c_str(),0,mid2.length()); cout << valid << endl; //输出false return 0;}