LeetCode102 Binary Tree level Order traversal Java

Exercises

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    /   7

Return its level order traversal as:

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

Answer:

Sequence traversal, a layer of nodes exist in a queue, and then sequentially traverse each node of the queue, if its child nodes are not empty continue to join the queue when the queue is empty traversal end

Code:

public static list<list<integer>> Levelorder (TreeNode root) {      list<list<integer>> result =new arraylist<list<integer>> ();//Returns the final result   queue<treenode> treenodes2=new LinkedList<> ();//the node used to store each layer      //Handle the first node (root)   if (root==null)   return result;   else {   list<integer> temp=new arraylist<> ();//Store temporary results   temp.add (root.val);   Treenodes2.offer (root);   Result.add (temp);     }    while (!treenodes2.isempty ())   {  int i=treenodes2.size ();  List<integer> tempreslut=new arraylist<> ()///store temporary results while  (i>0)//traverse all nodes of this layer  {    TreeNode Tnode=treenodes2.poll ();  if (tnode.left!=null)  {  tempreslut.add (tNode.left.val);  Treenodes2.offer (Tnode.left);  }  if (tnode.right!=null)  {  tempreslut.add (tNode.right.val);  Treenodes2.offer (tnode.right);  }  i--;  }  if (!tempreslut.isempty ()) Result.add (Tempreslut);     }      return result;    }

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LeetCode102 Binary Tree level Order traversal Java