根据树的后序判断是不是二叉搜索树&&二叉树中和为某一值的路径

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根据树的后序判断是不是二叉搜索树。

关键树的后序序列规律是根是最后的，而右子树都大于根，左子树都小于根。可以判断序列是否满足这个条件，然后递归判断左右子树即可。

#include<iostream>#include<string>using namespace std;bool IsSearchBinaryTree(int *numbers,int begin,int end){if(numbers == NULL || begin>end)return false;if(begin == end)return true;int root = numbers[end];int i = end-1;bool bleft,bright;//右子树都大于根值while(i>=begin && numbers[i] > root)i--;int j = i;//左子树都小于根值while(j>=begin && numbers[j] <root)j--;//如果还有结点剩余则不满足搜索二叉树性质if(j >=begin)return false;//判断左右子树是否为搜索二叉树bleft = bright = true;if(i >= begin)bleft = IsSearchBinaryTree(numbers,begin,i);if(i <end-1)bright= IsSearchBinaryTree(numbers,i+1,end-1);return (bleft && bright);}int main(){bool test1,test2;int number1[] = {5,7,6,9,11,10,8};test1 = IsSearchBinaryTree(number1,0,6);if(test1)printf("number1 is binary search sequence\n");elseprintf("number1 is not binary search sequence\n");int number2[] = {2,3,5,4,1,2,8};test2 = IsSearchBinaryTree(number2,0,6);if(test2)printf("number2 is binary search sequence\n");elseprintf("number2 is not binaru search sequence\n");system("PAUSE");return 0;}

路径指的是从根到叶子结点。可以用dfs。

#include<iostream>#include<string>#include<queue>using namespace std;struct Binarytree{      int value;      Binarytree *left;      Binarytree *right;  };  int path[20];Binarytree* Buildtree(){      int x;      scanf("%d",&x);      if(x == 0)          return NULL;      queue<Binarytree *> Bq;      Binarytree* root = (Binarytree *)malloc(sizeof(Binarytree));      root->value = x;      root->left = NULL;      root->right = NULL;      Binarytree* temp = root;      Bq.push(temp);      while(!Bq.empty()){          temp = Bq.front();          Bq.pop();          if(temp->left == NULL){              int x;              printf("输入%d的左结点\n",temp->value);              scanf("%d",&x);              if(x!=0){                  Binarytree * tleft = (Binarytree *)malloc(sizeof(Binarytree));                  tleft->value = x;                  tleft->left = tleft->right = NULL;                  temp->left = tleft;                  Bq.push(tleft);              }          }          if(temp->right == NULL){              int x;              printf("输入%d的右结点\n",temp->value);              scanf("%d",&x);              if(x!=0){                  Binarytree * tright = (Binarytree *)malloc(sizeof(Binarytree));                  tright->value = x;                  tright->left = tright->right = NULL;                  temp->right = tright;                  Bq.push(tright);              }          }      }      return root;  }  //sum表示要求的路径值，current和path用来记录路径void printtree(int sum,int current,int path[],Binarytree* root){if(sum <0)return;path[current] = root->value;++current;sum -= root->value;if(sum == 0 && root->left==NULL && root->right == NULL){printf("find a path:\n");for(int i =0;i<current;i++)printf("%d ",path[i]);printf("\n");return;}if(root->left)printtree(sum,current,path,root->left);if(root->right)printtree(sum,current,path,root->right);}int main(){Binarytree *root = Buildtree();printtree(22,0,path,root);system("PAUSE");return 0;}

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