# LeetCode题解

### 1、Merge Two Binary Trees

Given two binary trees and imagine that when you put one of them to cover the other, some nodes of the two trees are overlapped while the others are not.
You need to merge them into a new binary tree. The merge rule is that if two nodes overlap, then sum node values up as the new value of the merged node. Otherwise, the NOT null node will be used as the node of new tree.

``````Example 1:
Input:
Tree 1                     Tree 2
1                         2
/ \                       / \
3   2                     1   3
/                           \   \
5                             4   7
Output:
Merged tree:
3
/ \
4   5
/ \   \
5   4   7
``````

Note: The merging process must start from the root nodes of both trees.

``````/**
* Definition for a binary tree node.
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
class Solution {
public TreeNode mergeTrees(TreeNode t1, TreeNode t2) {
if(t1 != null && t2 != null) {
t1.left = mergeTrees(t1.left,t2.left);
t1.right = mergeTrees(t1.right,t2.right);
t1.val += t2.val;
return t1;
}
return t1 == null ? t2 : t1;
}
}
``````

### 2、leetcode Counting Bits

leetcode Counting Bits
Given a non negative integer number num. For every numbers i in the range 0 ≤ i ≤ num calculate the number of 1’s in their binary representation and return them as an array.
Example:
For num = 5 you should return [0,1,1,2,1,2].
It is very easy to come up with a solution with run time O(n*sizeof(integer)). But can you do it in linear time O(n) /possibly in a single pass?
Space complexity should be O(n).
Can you do it like a boss? Do it without using any builtin function like __builtin_popcount in c++ or in any other language.

#### 法1：仔细分析一下可发现当一个数是2的整数幂的时候，二进制1的个数为1。之后就是前一个序列+1，如：

1、2(0010) = 1
2、3(0011) = 2(0010) + 1(0001) = 2
3、6(0110) = 4(0100) + 2(0010) = 1 + 1 = 2
4、7(0111) = 4(0100) + 3(0011) = 1 + 2 = 3
5、9 (1001) = 8(1000) + 1(0001) = 1 + 1 = 2

``````public static int[] countBits(int num) {
int[] res = new int[num + 1];
int pow2 = 1,before = 1;
for(int i = 1;i<=num;i++){
if(pow2 == i) {
res[i] = before = 1;
pow2 <<= 1;
} else {
res[i] = res[before] + 1;
before++;
}
}
return res;
}
``````

#### 法2：一个数*2就是把它的二进制全部左移1位，也就是说1的个数是相等的，那我我们如果将一个数右移以为呢？如果这个数是偶数，右移1位正好是num/2,如果是奇数，则是num/2 + 1。我们可以得到以下公式：

res[i] = res[i >> 1] + (i & 1);(如果是奇数，i & 1得到1，如果是偶数i & 1得到0)

``````public static int[] countBits2(int num) {
int[] res = new int[num + 1];
for (int i = 1; i <= num; i++) {
res[i] = res[i >> 1] + (i & 1);
}
return  res;
}
``````

### 推荐阅读更多精彩内容

• 背景 一年多以前我在知乎上答了有关LeetCode的问题, 分享了一些自己做题目的经验。 张土汪：刷leetcod...
• 我究竟能走多远？ 时间过得真快，一眨眼，大一过去了，貌似一眨眼的功夫，干了好多事，体验了好多第一次。 去了南京，爬...
• 我认为一个成功的人，应该具有的两特点：事不宜迟、和反馈。在最近的生活中，我越来越感受到这两点的重要性。如果要和具体...
• 这是闪闪发光的词，在这两年里边，教你提升认知，教你如何思考，可以卖上好的价钱，这些课程的核心就是大家的思维能力太差...
• 有时听一首歌会想起一个人。 所以我几乎不听薛之谦的歌，因为我觉得林逸明的声音莫名的像他。我觉得薛之谦会让我想起以前...