leetcode-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.

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思路：递归，分解成左右子树是否有路径的和为sum-roor->val.

递归结束条件为当前节点值为sum且为叶子节点

/**
* 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) {
//递归,注意节点值可能为负数
if(root == NULL)
{
return false;
}
if(root->val == sum && root->left == NULL && root->right == NULL)
{
return true;
}

return hasPathSum(root->left,sum - root->val) || hasPathSum(root->right,sum - root->val);
}
};