404. Sum of Left Leaves
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
From: LeetCode
Link: 404. Sum of Left Leaves
Solution:
Ideas:
1. isLeaf: A helper function that checks if a node is a leaf (no left or right children).
2. sumOfLeftLeavesHelper: This is the main recursive function. It takes two arguments: the current node and a flag isLeft that indicates whether the node is a left child.
- If the node is a leaf and isLeft is true, it returns the node’s value.
- Otherwise, it recursively checks the left and right subtrees, summing the results.
3. sumOfLeftLeaves: This is the main function that starts the recursion, passing 0 as the initial isLeft flag (since the root is not a left child).
Code:
/*** Definition for a binary tree node.* struct TreeNode {* int val;* struct TreeNode *left;* struct TreeNode *right;* };*/int isLeaf(struct TreeNode* node) {return (node != NULL && node->left == NULL && node->right == NULL);
}int sumOfLeftLeavesHelper(struct TreeNode* root, int isLeft) {if (root == NULL) return 0;if (isLeaf(root) && isLeft) {return root->val;}// Recurse on left and right childrenreturn sumOfLeftLeavesHelper(root->left, 1) + sumOfLeftLeavesHelper(root->right, 0);
}int sumOfLeftLeaves(struct TreeNode* root) {return sumOfLeftLeavesHelper(root, 0);
}
