404. Sum of Left Leaves (Easy)

Given the root of a binary tree, return the sum of all left leaves.

A leaf is a node with no children. A left leaf is a leaf that is the left child of another node.

 

Example 1:

Input: root = [3,9,20,null,null,15,7]
Output: 24
Explanation: There are two left leaves in the binary tree, with values 9 and 15 respectively.

Example 2:

Input: root = [1]
Output: 0

 

Constraints:

• The number of nodes in the tree is in the range [1, 1000].
• -1000 <= Node.val <= 1000

A:

/**

* Definition for a binary tree node.

* struct TreeNode {

* int val;

* TreeNode *left;

* TreeNode *right;

* TreeNode() : val(0), left(nullptr), right(nullptr) {}

* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}

* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}

* };

*/

class Solution {

public:

int sumOfLeftLeaves(TreeNode* root) {

return helper(root, false);

}

private:

int helper(TreeNode* root, bool isLeftChild){

if(!root)

return 0;

if(!root->left && !root->right && isLeftChild)

return root->val;

return helper(root->left, true) + helper(root->right, false);

}

};

NOTE：
如果只有一个根节点，  返回0