Binary Tree Zigzag level Order traversal

Binary Tree Zigzag level Order traversal

Problem:

Given a binary tree, return the zigzag level order traversal of its nodes ' values. (ie, from left-to-right, then right-to-left for the next level and alternate between).

Ideas:

Queue Hierarchy Access

My Code:

 Public classSolution { PublicList<list<integer>>Zigzaglevelorder (TreeNode root) {List<List<Integer>> rst =NewArraylist<list<integer>>(); if(Root = =NULL)returnrst; Queue<TreeNode> queue =NewLinkedlist<treenode>();        Queue.offer (root); intCount = 1;  while(!Queue.isempty ()) {            intSize =queue.size (); List<Integer> list =NewArraylist<integer>();  for(inti = 0; i < size; i++) {TreeNode node=Queue.poll (); if(Node.left! =NULL) Queue.offer (node.left); if(Node.right! =NULL) Queue.offer (node.right);            List.add (Node.val); }            if(count%2 = = 1) {rst.add (list); }            Else{collections.reverse (list);            Rst.add (list); } Count++; }        returnrst; }}

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Others code:

 Public classSolution { PublicArraylist<arraylist<integer>>Zigzaglevelorder (TreeNode root) {ArrayList<ArrayList<Integer>> result =NewArraylist<arraylist<integer>>(); if(Root = =NULL) {            returnresult; } Stack<TreeNode> Currlevel =NewStack<treenode>(); Stack<TreeNode> Nextlevel =NewStack<treenode>(); Stack<TreeNode>tmp;        Currlevel.push (root); BooleanNormalorder =true;  while(!Currlevel.isempty ()) {ArrayList<Integer> Currlevelresult =NewArraylist<integer>();  while(!Currlevel.isempty ()) {TreeNode node=Currlevel.pop ();                Currlevelresult.add (Node.val); if(normalorder) {if(Node.left! =NULL) {Nextlevel.push (node.left); }                    if(Node.right! =NULL) {Nextlevel.push (node.right); }                } Else {                    if(Node.right! =NULL) {Nextlevel.push (node.right); }                    if(Node.left! =NULL) {Nextlevel.push (node.left);            }}} result.add (Currlevelresult); TMP=Currlevel; Currlevel=Nextlevel; Nextlevel=tmp; Normalorder= !Normalorder; }        returnresult; }}

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The Learning Place:

• Although the use of queues can also achieve this problem, but think about, if this layer of the right node is the last element of this layer, then it is the next level of the last access, if the leftmost node of this layer is the last element of the layer access, then the left of the element will be the next layer of access to the node, Therefore, it is more reasonable to use a stack than a queue, and this can save the time of the list rollover under the queue method.

Binary Tree Zigzag level Order traversal