以 A0 作为起点,对下面的无向图进行深度优先遍历时,遍历顺序不可能是
const int TREE_SIZE = 9; std::stack<node*> visited, unvisited; node nodes[TREE_SIZE]; node* current; for( int i=0; i<TREE_SIZE; i++) //初始化树 { nodes[i].self = i; int child = i*2+1; if( child<TREE_SIZE ) //Left child nodes[i].left = &nodes[child]; else nodes[i].left = NULL; child++; if( child<TREE_SIZE ) //Right child nodes[i].right = &nodes[child]; else nodes[i].right = NULL; } unvisited.push(&nodes[0]); //先把0放入UNVISITED stack while(!unvisited.empty()) //只有UNVISITED不空 { current=(unvisited.top()); //当前应该访问的 unvisited.pop(); if(current->right!=NULL) unvisited.push(current->right); // 把右边压入 因为右边的访问次序是在左边之后 if(current->left!=NULL) unvisited.push(current->left); visited.push(current); cout<<current->self<<endl; }