190. Reverse Bits

Reverse bits of a given 32 bits unsigned integer.
For example, given input 43261596 (represented in binary as 00000010100101000001111010011100), return 964176192 (represented in binary as 00111001011110000010100101000000).

看題目大概就知道要做什麼了

class Solution {
public:
    uint32_t reverseBits(uint32_t n) {
        uint32_t x;
        for(auto i = 31; n; ) {
            x |= (n & 1) << i;
            n >>= 1;
            -- i;
        }
        return x;
    }
};

110. Balanced Binary Tree

Given a binary tree, determine if it is height-balanced.
For this problem, a height-balanced binary tree is defined as a binary tree in which the depth of the two subtrees of every node never differ by more than 1.

檢查這個一顆二元樹是不是高度平衡(這邊的定義為左右兩邊的子樹高度不可以相差大於1)

解法

看code就清楚了

/**
 * 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:  
        bool isBalanced(TreeNode* root) 
        {  
            return maxDepth(root) != -1;
        }  
          
        int maxDepth(TreeNode* node)
        {  
            if(!node) 
            {
                return 0;
            }
            
            auto ldepth = maxDepth(node->left);
            auto rdepth = maxDepth(node->right);
            
            if(ldepth == -1 || rdepth == -1 || abs(ldepth - rdepth) > 1)
            {
                return -1;
            }
            
            return max(ldepth, rdepth) + 1;  
        }  
    };  

113. Path Sum II

Given a binary tree and a sum, find all root-to-leaf paths where each path's sum equals the given sum.
For example:
Given the below binary tree and sum = 22,

              5
             / \
            4   8
           /   / \
          11  13  4
         /  \    / \
        7    2  5   1

return

[
   [5,4,11,2],
   [5,8,4,5]
] 

簡單來說，就是要你找出所有符合條件的路徑。 

解法也沒有很難，跟78. Subsets的邏輯是差不多的

/**
 * 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:
    vector<vector<int>> pathSum(TreeNode* root, int sum) {
        
        vector<vector<int>> answer;
        vector<int> output;
        
        if (root == nullptr)
        {
            return answer;   
        }
        
        PathSumDFS(root, sum, answer, output);         
        return answer; 
                     
    }
    
    void PathSumDFS(TreeNode* node, int sum, vector<vector<int>> &answer, vector<int> &output)
    {         
        if (node == nullptr)
        {
            return;
        }
        
        output.push_back(node->val);
        
        if (node->left == nullptr && node->right == nullptr)
        {
            if(node->val == sum)
            {
                answer.push_back(output);
            }
        }
        
        PathSumDFS(node->left, sum - node->val, answer, output); 
        PathSumDFS(node->right, sum - node->val, answer, output);
        //返回上一層時需要把這個element清除
        output.pop_back();
        
    }    
};