Binary Tree Level Order Traversal leetcode

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

For example:
Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

[
  [3],
  [9,20],
  [15,7]
]

这道题是广度优先搜索的模板，BFS用queue来实现
时间O(n) 空间O(n)

要注意：a.有while和for双重循环 
b.每次都要给size付一个新值（如果不赋值queue.size在不停变化）
c.queue add 和 delete 是.offer 和.poll 
d. queue为什么用linkedlist实现？？？---Queue是接口， LinkedList可以实现此接口。

public class Solution {
    public List<List<Integer>> levelOrder(TreeNode root) {
        List<List<Integer>> res= new ArrayList<List<Integer>>();
        if (root == null) {
            return res;
        }
        Queue<TreeNode> queue = new LinkedList<TreeNode>();
        queue.offer(root);
        while (!queue.isEmpty() ) {
            int size = queue.size();
            List<Integer> tem = new ArrayList<Integer>();
            for (int i = 0; i < size; i++) {
                TreeNode node = queue.poll();
                tem.add(node.val);
                if (node.left != null) {
                    queue.offer(node.left);
                }
                if (node.right != null) {
                    queue.offer(node.right);
                }
            }
            res.add(tem);
        }
        return res;
    }
}

Binary Search

For a given sorted array (ascending order) and a target number, find the first index of this number in O(log n) time complexity.

If the target number does not exist in the array, return -1.

Example

If the array is [1, 2, 3, 3, 4, 5, 10], for given target 3, return 2.

这是一个经典的binary serch的模板

 1.start+1 < end

2. mid = start + (end-start/)2

3. nums[mid] <, ==,> target 的三种情况

4. 是return start end 还是-1

class Solution {
    /**
     * @param nums: The integer array.
     * @param target: Target to find.
     * @return: The first position of target. Position starts from 0.
     */
    public int binarySearch(int[] nums, int target) {
        if (nums.length == 0){
            return -1;
        }
        int start = 0;
        int end = nums.length - 1;
        int mid;
        while (start + 1 < end){
            mid = start + (end - start) / 2;
            if (target > nums[mid]){
                start = mid;
            } else if (target < nums[mid]) {
                end = mid;
            } else {
                end = mid;
            }
        }
        if (nums[start] == target){
            return start;
        } else if (nums[end] == target){
            return end;
        } else {
            return -1;
        }
    }
}