1st round, 314 binaryTreeVerticalTraversal

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<List<Integer>> verticalOrder(TreeNode root) {        List<List<Integer>> result = new ArrayList<>();        if(root==null) return result;                Queue<TreeNode> nodeQ = new LinkedList<>();  // have a clear idea of what nodeQ contains.        Queue<Integer> levelQ = new LinkedList<>();  // have a clear idea of what levelQ contains.        Map<Integer, List<Integer>> map = new HashMap<>(); // have a clear idea of what map contains.                nodeQ.add(root);        levelQ.add(0);                int min=0;        int max=0;                while(!nodeQ.isEmpty()){   // while(nodeQ!=null) is wrong!! should check whether it is empty.            TreeNode node=nodeQ.poll();            int level=levelQ.poll();                        if(!map.containsKey(level)) map.put(level, new ArrayList<Integer>());            map.get(level).add(node.val); // have a clear idea of what map contains.                        if(node.left!=null){                nodeQ.add(node.left);                levelQ.add(level-1);                min=Math.min(min, level-1);            }                        if(node.right!=null){                nodeQ.add(node.right);                levelQ.add(level+1);                max=Math.max(max, level+1);            }            // for the above two if() statements only if we have some in either left or right child, we add the child to the Q, otherwise need to do nothing.        }                for(int i=min; i<=max; i++){            result.add(map.get(i));        }        return result;    }}        // have no idea about the question in the beginning, but after looking through solutions I immediately get the idea, it is the BFS approach. Did I know BFS?? Of course!!! You did many quesitons on BFS, basically it is reading elems on same level of tree, and then move to the next level.        // here there are many things for me to 总结, such as using Queue<> (Interface) and LinkedList<>(class) to track the information on each level. ALSO, understanding this question and transfering it to some simple questions and easy solutions is a good practice. Lastly, knowing using min, max to keep record of the range is the key point to this question!! 一会好好总结在ppt上。。        

这道题，虽然代码略多，但是思路还是非常清晰简练的。。。相信读者看代码就能知道我什么意思。。