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112. 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.
一开始被这题目坑了，因为叶子节点的左右孩子都是NULL 后捞改了一下代码 过了
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: bool hasPathSum1(TreeNode* root, int sum) { if (root == NULL){ if (sum == 0) return true; return false; } if (root->left == NULL) return hasPathSum1(root->right,sum - root->val); else if (root->right == NULL) return hasPathSum1(root->left,sum - root->val); else return (hasPathSum1(root->left,sum - root->val) || hasPathSum1(root->right,sum - root->val)); } bool hasPathSum(TreeNode* root, int sum) { if (root == NULL) return false; return hasPathSum1(root, sum); }};
/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode(int x) : val(x), left(NULL), right(NULL) {} * }; */class Solution {public: bool hasPathSum(TreeNode* root, int sum) { if(!root) return false; if(sum == root->val && root->left==NULL &&root->right == NULL) return true; return hasPathSum(root->left, sum-root->val)||hasPathSum(root->right, sum-root->val); }};
112. Path Sum
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