Leetcode: Binary Tree Level Order Transversal

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]
]

Best approach:

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

把树看成一个有向图，进行BFS。BST的通常做法都是维护一个队列。这道题的难度在于，如何在队列不断的入队出队操作中，确定哪些node是一个层次的。想了很久没什么好办法，参考了网上的做法，发现他在维护两个数：一个是父节点所在层次的节点在当前队列中的数目ParentNumInQ，另一个是子节点所在层次的节点在当前队列中的数目ChildNumInQ。初始数值为1和0. 每次父节点出队，ParentNumInQ--，每次子节点入队，ParentNumInQ++。一旦ParentNumInQ减至0，说明当前父节点层次已全部出队，全部被存入LinkedList中，这就得到了一层的所有节点。接下来需要一些更新，ParentNumInQ = ChildNumInQ, ChildNumInQ = 0开始考虑下一层。

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public ArrayList<ArrayList<Integer>> levelOrder(TreeNode root) {
         ArrayList<ArrayList<Integer>> lists = new ArrayList<ArrayList<Integer>> ();
         if (root == null) return lists;
         LinkedList<TreeNode> queue = new LinkedList<TreeNode>();
         queue.add(root);
         int ParentNumInQ = 1;
         int ChildNumInQ = 0;
         ArrayList<Integer> list = new ArrayList<Integer>();
         while (!queue.isEmpty()) {
             TreeNode cur = queue.poll();
             list.add(cur.val);
             ParentNumInQ--;
             if (cur.left != null) {
                 queue.add(cur.left);
                 ChildNumInQ++;
             }
             if (cur.right != null) {
                 queue.add(cur.right);
                 ChildNumInQ++;
             }
             if (ParentNumInQ == 0) {
                 ParentNumInQ = ChildNumInQ;
                 ChildNumInQ = 0;
                 lists.add(list);
                 list = new ArrayList<Integer>();
             }
         }
         return lists;
     }
 }