[LeetCode 102] Level-by-Level Traversal (medium)

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,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]

Solution

• 只要是与Tree和Level相关的问题，大部分都可以用BFS。

• BFS的套路： 用2个Queue 或者Stack。第一个用于存储当前正在扫描层 , 第二个用于存储下一层Object.

  1. 先把root放入queue1，防止queue1为空。
  2. 当queue1 != null: 遍历queue1中每个节点，将其value放入subArray (每一层的结果)；同时将其子节点放入queue2。
  3. 当queue1 == null:
     ---- 将subArray加入到最终结果中，清空subArray；
     ---- 然后换层：queue1 = queue2; queue2 = new LinkedList<> ()

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public List> levelOrder(TreeNode root) {
        List> result = new ArrayList<> ();
        if (root == null) {
            return result;
        }
        
        Queue  queue1 = new LinkedList<> ();
        Queue  queue2 = new LinkedList<> ();
        queue1.add (root);
        
        // subResult for each layer
        List subArray = new ArrayList<> ();
        
        while (!queue1.isEmpty ()) {
            TreeNode current = queue1.poll ();
            subArray.add (current.val);
            
            if (current.left != null) {
                queue2.add (current.left);
            }
            
            if (current.right != null) {
                queue2.add (current.right);
            }
            
            // queue1 is empty and then swap queue1 and queue2, and add subResult into result
            if (queue1.isEmpty ()) {
                result.add (subArray);
                subArray = new ArrayList ();
                
                queue1 = queue2;
                queue2 = new LinkedList ();
            }
            
        }
        
        return result;
    }
}