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Sum of Left Leaves
Find the sum of all left leaves in a given binary tree.
Example:
3 / 9 20 / 15 7There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.
Analyse: Check if the current node has left child and if the left child is a leaf node.
Runtime: 3ms
1 /** 2 * Definition for a binary tree node. 3 * struct TreeNode { 4 * int val; 5 * TreeNode *left; 6 * TreeNode *right; 7 * TreeNode(int x) : val(x), left(NULL), right(NULL) {} 8 * }; 9 */10 class Solution {11 public:12 int sumOfLeftLeaves(TreeNode* root) {13 if (!root || (!root->left && !root->right)) return 0;14 15 int result = 0;16 sumLeftLeaves(root, result);17 return result;18 }19 20 void sumLeftLeaves(TreeNode* root, int &result) {21 if (root->left && !root->left->left && !root->left->right)22 result += root->left->val;23 24 if (root->left) sumLeftLeaves(root->left, result);25 if (root->right) sumLeftLeaves(root->right, result);26 }27 };
Sum of Left Leaves
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