[Leetcode] Binary Tree level Order traversal II

Question:

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

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

    3   /   9    /   7

Return its bottom-up level order traversal as:

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

1. Classification of types of questions:

2, the idea: first calculate the depth, recursively add each point, according to the depth, determine where the linked list

3. Complexity of Time:

4. Code:

 Public classSolution { PublicList<list<integer>>levelorderbottom (TreeNode root) {List<List<Integer>> list=NewArraylist<list<integer>>(); if(root==NULL)returnlist; intDepth=depth (root);  for(inti=0;i<depth;i++) {List<Integer> temp=NewArraylist<integer>();        List.add (temp); } traverdepth (Root, List,1, depth);        Collections.reverse (list); returnlist; }     Public voidTraverdepth (TreeNode node,list<list<integer>> List,intCurdepth,intdepth) {                if(node!=NULL{List.get (curDepth-1). Add (Node.val);}Else return; Curdepth++; if(node.left!=NULL) traverdepth (Node.left, list, curdepth, depth); if(node.right!=NULL) traverdepth (node.right, list, curdepth, depth); }     Public intdepth (TreeNode root) {if(root==NULL){              return0; }          return1 +Math.max (Depth (root.left), Depth (root.right)); } }

5. Optimization:

　　Some methods on the Internet use two auxiliary linked list, the first record when the layer of node, the second record all the next layer of node, after traversing, will be the first to return to the total linked list, when the layer points to the next layer, the next layer is empty, so continue to iterate down.

　　

List<list<integer>> list=NewArraylist<list<integer>>(); if(root==NULL)returnlist; List <TreeNode> listcur=new arraylist<treenode>(); List<TreeNode> listnex=new arraylist<treenode>(); Listcur.add (root);//Add First         while(!Listcur.isempty ()) {List<Integer> listtemp=NewArraylist<integer>();  for(intI=0;i<listcur.size (); i++) {TreeNode TreeNode=Listcur.get (i);                Listtemp.add (Treenode.val); if(treenode.left!=NULL) Listnex.add (treenode.left); if(treenode.right!=NULL) Listnex.add (treenode.right);                                    } list.add (Listtemp); List<TreeNode> temp=listcur; Listcur=Listnex; Listnex=temp;        Listnex.clear ();        } collections.reverse (list); returnList

6. Expansion

　　Order

 Public classSolution { PublicList<list<integer>>Levelorder (TreeNode root) {List<List<Integer>> list=NewArraylist<list<integer>>(); if(root==NULL)returnlist; List<TreeNode> listcur=NewArraylist<treenode>(); List<TreeNode> listnex=NewArraylist<treenode>(); Listcur.add (root);//Add First         while(!Listcur.isempty ()) {List<Integer> listtemp=NewArraylist<integer>();  for(intI=0;i<listcur.size (); i++) {TreeNode TreeNode=Listcur.get (i);                Listtemp.add (Treenode.val); if(treenode.left!=NULL) Listnex.add (treenode.left); if(treenode.right!=NULL) Listnex.add (treenode.right);                                    } list.add (Listtemp); List<TreeNode> temp=listcur; Listcur=Listnex; Listnex=temp;        Listnex.clear (); }        returnlist; }}

[Leetcode] Binary Tree level Order traversal II