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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: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,考虑root为NULL的情况,异常判断;
2,结构体指针后面应该接->,而不是点。因为是指针。
/**
* Definition for binary tree * 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(NULL == root) return false; if((sum == root->val)&&(NULL == root->left)&&(NULL == root->right)) { return true; } if(NULL != root->left) root->left->val = root->left->val + root->val; if(NULL != root->right) root->right->val = root->right->val + root->val; if(hasPathSum(root->left,sum)||hasPathSum(root->right,sum)) { return true; } else return false; } };转载地址:http://dpbci.baihongyu.com/