Sum of Left Leaves

Find the sum of all left leaves in a given binary tree.

Example:

    3
   / \
  9  20
    /  \
   15   7

There are two left leaves in the binary tree, with values 9 and 15 respectively. Return 24.

/**
 * 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:
    int sumOfLeftLeaves(TreeNode* root) {
        int ans = 0;
        TreeNode *prev = NULL;
        helper(root, NULL, ans);
        return ans;
    }
   
    void helper(TreeNode* root, TreeNode *prev, int& ans) {
        if (root == NULL) {
            return;
        }
        if (prev == NULL) {
            helper(root -> left, root, ans);
            helper(root->right, NULL, ans);
            return;
        }
        if (root -> left == NULL && root -> right == NULL && root == prev -> left) {
            ans += root -> val;
        }
        prev = root;
        helper(root -> left, prev, ans);
        helper(root -> right, NULL, ans);
        return;
    }
};

==== Simpler ======

class Solution { private: int res=0; void dfs(TreeNode* node, bool isLeft) { // root为叶子的情况 if(node->left == NULL && node->right == NULL) { if(isLeft) res+=node->val; return; } if(node->left!=NULL) dfs(node->left, true); if(node->right!=NULL) dfs(node->right, false); } public: int sumOfLeftLeaves(TreeNode* root) { if(!root) return 0; dfs(root, false); return res; } };