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.
//利用深度优先搜索DFS,当搜索到的节点为叶子节点时。才推断总和是否等于给定的sum值 class Solution { public: bool dfs(TreeNode* node, int sum, int curSum){ if (node == NULL) return false; if (node->left == NULL && node->right == NULL) return curSum + node->val == sum; return dfs(node->left, sum, curSum + node->val) || dfs(node->right, sum, curSum + node->val); } bool hasPathSum(TreeNode* root, int sum) { return dfs(root, sum, 0); } };