难度:Easy
原题连接
内容描述
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.
Note: A leaf is a node with no children.
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.
> 思路
******- 时间复杂度: O(N)******- 空间复杂度: O(1)******
DFS遍历二叉树,分别计算左子树和右子树的是否存在 sum - val 的路径
```cpp
/**
* 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 == nullptr)
return false;
if(root ->right == nullptr && root ->left == nullptr)
return sum - root ->val ? false : true;
return hasPathSum(root ->left,sum - root ->val) || hasPathSum(root ->right,sum - root ->val);
}
};