ARTS-week45

Algorithms

Sum of Left Leaves
思路: 二叉树遍历，求左子叶和的话，需要知道当前结点是否是左子节点，如果是左子节点，而且该左子节点不存在子节点，说明其是左子叶，就将其值加入结果sum中。

/**
 * 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) {
        if (!root || (!root->left && !root->right)) {
            return 0;
        }
        int sum = 0;
        helper(root, sum);
        return sum;
    }

    void helper(TreeNode* node, int& sum) {
        if (node->left)
            helper(node->left, sum);
        if (node->left && !node->left->left && !node->left->right)
            sum += node->left->val;
        if (node->right) 
            helper(node->right, sum);
    }
};

Convert a Number to Hexadecimal
思路: 先将十进制转换成二进制，再将二进制每4位转成一个十六进制字符。十进制转二进制时，只需要num & 15(0xf)，再将num右移四位，要注意题目中有说The given number is guaranteed to fit within the range of a 32-bit signed integer.，所以要注意次数，否则会报出Memory Limit Exceeded。

class Solution {
public:
    string toHex(int num) {
        if(num == 0) {
            return "0";
        }
        
        string result = "";
        const string s = "0123456789abcdef";
        
        while (num != 0 && result.length() < 8) {
            result += s[num & 0xf];
            num >>= 4;
        }
        reverse(result.begin(), result.end());
        return result;
    }
};