题目大意：一颗二叉树，每个节点都有一个Value, 判断根节点到叶节点的路径求和值是否等于某个数Sum.
比如说如下这样一颗二叉树，76是45，21,10这条路径的求和值，77就没有满足条件的路径。
45
21 65
10 24 50 70
代码依旧用C++来实现，二叉树一般采用递归的方式来解决。
1 #include <iostream> 2 3 using namespace std; 4 5 typedef struct BTree 6 7 { 8 int value; 9 struct BTree* left; 10 struct BTree* right; 11 } BTree; 12 typedef struct BTree Node; 13 14 //recursively insert a tree node 15 16 BTree* Insert(BTree* T, int value) 17 { 18 if(T == NULL) 19 { 20 T = (BTree*)malloc(sizeof(struct BTree)); 21 if(T == NULL) 22 printf("Malloc failed"); 23 else{ 24 T->value = value; 25 T->left = T->right = NULL; 26 } 27 } 28 else if(value < T->value) 29 T->left = Insert(T->left,value); 30 else if(value > T->value) 31 T->right = Insert(T->right,value); 32 return T; 33 } 34 BTree* MakeEmpty(BTree* T) 35 { 36 if(T != NULL){ 37 MakeEmpty(T->left); 38 MakeEmpty(T->right); 39 free(T); 40 } 41 return NULL; 42 } 43 44 45 bool hasPathSum( Node *node, int sum, int pathSum) 46 { 47 bool match = false; 48 if(node != NULL) 49 pathSum += node->value; 50 if(node->left== NULL && node->right == NULL) 51 { 52 if(sum == pathSum) 53 match = true; 54 else 55 match = false; 56 } 57 if(node->left != NULL && !match) 58 match = hasPathSum(node->left,sum,pathSum); 59 if(node->right != NULL && !match) 60 match = hasPathSum(node->right,sum,pathSum); 61 return match; 62 } 63 bool hasPathSum( Node *root, int sum) 64 { 65 if(root == NULL) return false; 66 bool match = false; 67 match = hasPathSum(root,sum,0); 68 return match; 69 } 70 int main() 71 { 72 BTree* T = NULL; 73 T = Insert(T,45); 74 T = Insert(T,21); 75 T = Insert(T,65); 76 T = Insert(T,10); 77 T = Insert(T,50); 78 T = Insert(T,70); 79 T = Insert(T,24); 80 bool match = hasPathSum(T,76); 81 cout << match << endl; 82 match = hasPathSum(T,77); 83 cout << match << endl; 84 MakeEmpty(T); 85 return 0; 86 }