Problem:

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) {}
 * };
 */

Solution:

Recursive solution:

檢查當前節點的左子節點是否是左子葉，如果是的話，返回左子葉的值加上對當前結點的右子節點調用遞歸的結果；如果不是的話，對左右子節點分別調用遞歸函數，返回二者之和。

class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (root == NULL) return 0;
        if (root->left && !root->left->left && !root->left->right) 
        {   //if left child is a leaf
            return root->left->val + sumOfLeftLeaves(root->right);
        }
        else return sumOfLeftLeaves(root->left) + sumOfLeftLeaves(root->right);
    }
};

Iterative solution:

class Solution {
public:
    int sumOfLeftLeaves(TreeNode* root) {
        if (!root || (!root->left && !root->right)) return 0;
        int result = 0;
        queue<TreeNode*> q;
        q.push(root);
        while (!q.empty()) {
            TreeNode *t = q.front(); 
            q.pop();
            if (t->left && !t->left->left && !t->left->right) 
            {  //if left child is a leaf
               result += t->left->val;
            }
            if (t->left) q.push(t->left);
            if (t->right) q.push(t->right);
        }
        return result;
    }
};

Memo:

讀清題意，left leaves。
考慮清楚遞歸的分界，最小情況是什么。