首页 > 代码库 > Path Sum
Path Sum
Given a binary tree and a sum, determine if the tree has a root-to-leaf path such that adding up all the values along the path equals the given sum.
For example:
Given the below binary tree and sum = 22,
5 / 4 8 / / 11 13 4 / \ 7 2 1
return true, as there exist a root-to-leaf path 5->4->11->2 which sum is 22.
1 /** 2 * Definition for binary tree 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */10 class Solution {11 public:12 bool hasPathSum(TreeNode *root, int sum) {13 14 if(root == NULL)15 return false;16 17 int tempSum = 0;18 int flag = 0;19 pathSum(root, tempSum, sum, flag);20 if(flag == 1)21 return true;22 else23 return false;24 }25 26 void pathSum(TreeNode *root, int tempSum, int sum, int &flag)27 {28 if(root != NULL)29 {30 if(root->left == NULL && root->right == NULL)31 {32 tempSum += root->val;33 if(tempSum == sum)34 flag = 1;35 36 return;37 }38 39 else40 {41 tempSum += root->val;42 pathSum(root->left, tempSum, sum, flag);43 pathSum(root->right, tempSum, sum, flag);44 }45 }46 }47 };
Path Sum
声明：以上内容来自用户投稿及互联网公开渠道收集整理发布，本网站不拥有所有权，未作人工编辑处理，也不承担相关法律责任，若内容有误或涉及侵权可进行投诉： 投诉/举报 工作人员会在5个工作日内联系你，一经查实，本站将立刻删除涉嫌侵权内容。