404. Sum of Left Leaves 解题记录

题目描述：

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

例子：

 

    3
   / \
  9  20
    /  \
   15   7

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

 

解题思路：

用递推对二叉树进行遍历，判断是否为末枝的左子叶，然后将所有的末枝的左子叶相加（不要忘了考虑空指针的情况）

代码：

 

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */
int sumOfLeftLeaves(struct TreeNode* root) {
    if (!root)
    //输入为空指针时
    {
        return 0;
    }
    int leftLeavesSum = 0;
    if (root->left)
    {
        if (!root->left->left && !root->left->right)
        //结束的条件也就是末枝的左子叶时
        {
            leftLeavesSum += root->left->val;
        }
        else
        {
                leftLeavesSum += sumOfLeftLeaves(root->left);
            }
        }
    if (root->right)
        {
            leftLeavesSum += sumOfLeftLeaves(root->right);
        }
    return leftLeavesSum;
}

 

解题收获：

对于C语言链表的使用还是有些不够熟练，还需要多加练习。