UVa 122 - Trees on the level

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题目：给你一颗二叉树上面的若干节点上的值（均为正数），判断从根到所有的给定的点的路径上的节点，

是不是都有值，且只被赋值一次。

分析：递归，搜索，树。

首先，利用递归建树，过程中记录是否有节点赋值两次；

然后，利用递归判断，查找是否有未被赋值的节点；

最后，利用bfs安深度输出每层的值。

说明：强大的sscanf函数╮(╯▽╰)╭。

#include <algorithm>#include <iostream>#include <cstdlib>#include <cstring>#include <cstdio>#include <cmath>using namespace std;typedef struct tnode{int    V;tnode* L;tnode* R;}tree;tree node[300];tree*root;int  tree_size,complete;int test(tree *root){if (!root->V) return 0;int ans = 0;if (!root->L || test(root->L))ans ++;if (!root->R || test(root->R))ans ++;return ans == 2;}void madetree(tree *root, char* str, int v){if (*str == 'R') {if (!root->R)root->R = &node[++ tree_size];madetree(root->R, str+1, v);}else if (*str == 'L') {if (!root->L)root->L = &node[++ tree_size];madetree(root->L, str+1, v);}else {if (root->V)complete = 0;root->V = v;}}tree*Q[300];void output(tree *root){int move = 0,save = 0;Q[save ++] = root;printf("%d",root->V);while (move < save) {tree* now = Q[move ++];if (now->L) {printf(" %d",now->L->V);Q[save ++] = now->L;}if (now->R) {printf(" %d",now->R->V);Q[save ++] = now->R;}}printf("\n");}int main(){char buf[256],leaf[256];int  value;complete = 1;root = &node[tree_size = 0];while (~scanf("%s",buf)) {if (!strcmp(buf,"()")) {if (!complete || !test(root))printf("not complete\n");else output(root);memset(node, 0, sizeof(node));root = &node[tree_size = 0];complete = 1;}else {sscanf(buf,"(%d,%s",&value,leaf);madetree(root, leaf, value);}}    return 0;}

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