Description
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.
Explain
这道题一看就是深度优先搜索的问题了。题意很简单，就是从找一个条路，从根部到叶子，其中节点的值加起来等于给定的值。那就遍历每个节点，每次到达一个新的叶子节点时，加上当前结点的值，看是否等于给定的值，如果是，那么就找到了。如果不是，退回上一层继续遍历，直到遍历完整个树，如果还找不到，那就无解。
Solution
/**
* 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) {
bool flag = false;
if (!root) return false;
dfs(flag, root, sum, 0);
return flag;
}
void dfs(bool& flag, TreeNode* root, int sum, int _sum) {
if (flag) return;
if (root) {
dfs(flag, root->left, sum,_sum + root->val);
dfs(flag, root->right, sum,_sum + root->val);
_sum += root->val;
if(!root->left&&!root->right&&_sum == sum) flag = true;
}
}
};
