题目来源：https://leetcode.cn/problems/sum-of-left-leaves/description/

 C++题解1：递归法，前序遍历。

1. 确定输入参数：当前节点，左叶子的和；

2. 确定终止条件：空节点时返回；

3. 确定单层递归逻辑：当该节点有左子树，判断是否为叶子节点，是则把它加入sum中，不是则进行下一步递归；当该节点有右子树，则进行下一步递归。

/**
 * 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:
    void getleft(TreeNode* node, int& sum) {
        if(node == nullptr) return;
        if(node->left) {
            if(node->left->left == nullptr && node->left->right == nullptr) {
                sum = sum + node->left->val;
            }
            else getleft(node->left, sum);
        }
        if(node->right) getleft(node->right, sum);
        return;
    }
    int sumOfLeftLeaves(TreeNode* root) {
        int sum = 0;
        if(root == nullptr) return 0;
        getleft(root, sum);
        return sum;
    }
};

C++题解2：迭代法，层序遍历

class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        stack<TreeNode*> st;
        if (root == NULL) return 0;
        st.push(root);
        int result = 0;
        while (!st.empty()) {
            TreeNode* node = st.top();
            st.pop();
            if (node->left != NULL && node->left->left == NULL && node->left->right == NULL) {
                result += node->left->val;
            }
            if (node->right) st.push(node->right);
            if (node->left) st.push(node->left);
        }
        return result;
    }
};