Leetcode Binary Tree Zigzag level Order traversal

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).

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

    3   /   9    /   7

Return its zigzag level order traversal as:

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

Confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

OJ ' s Binary Tree serialization:

The serialization of a binary tree follows a level order traversal, where ' # ' signifies a path terminator where no node ex Ists below.

Here's an example:

   1  /  2   3    /   4         5

The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}" .

Test instructions: The hierarchy traverses a tree, flipping over odd layers.

Idea: or use the queue hierarchy traversal, more than one to determine whether to flip the tag on the line, there is a reference to Java is passed, so every time you want to re-declare an object, or ans inside is pointing to the same block of memory.

/** * Definition for Binary tree * public class TreeNode {* int val, * TreeNode left, * TreeNode right; TreeNode (int x) {val = x;} *} */public class Solution {public list<list<integer>> Zigzaglevelorder (Tre    Enode root) {list<list<integer>> ans = new arraylist<list<integer>> ();        if (root = null) return ans;    queue<treenode> queue = new linkedlist<treenode> ();    Queue.add (root);    Boolean reverse = false;    while (!queue.isempty ()) {list<integer> tmp = new arraylist<integer> ();    int num = Queue.size ();    for (int i = 0; i < num; i++) {TreeNode t = queue.poll ();    Tmp.add (T.val);    if (t.left! = null) Queue.add (t.left);    if (t.right! = null) Queue.add (t.right);    } if (reverse) {collections.reverse (TMP);    Reverse = false;    } else reverse = true;    Ans.add (TMP);     } return ans; }}

Leetcode Binary Tree Zigzag level Order traversal