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,null,null,15,7],

    3   / \  9  20    /  \   15   7

return its zigzag level order traversal as:

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

对树进行Z形的层次遍历。首先想到的是，把所有的节点放入链表中，当行数为奇数行时，对该链表的节点进行反转。

public class Solution {    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {        int cnt = 0;int f = 0;int t = 1;int index; List<List<Integer>> lsts = new ArrayList<List<Integer>>();List<Integer> lst = new ArrayList<Integer>();List<TreeNode> nodeList = new ArrayList<TreeNode>();nodeList.add(root);if(root == null){return lsts;}index = 0;while(t != 0){            if(f == t){            t = cnt;            f = 0;            cnt = 0;            lsts.add(lst);            lst = new ArrayList<Integer>();            }            else{            TreeNode node = nodeList.get(index ++);            if(node != null){            f ++;            lst.add(node.val);            }            if(node.left != null){            cnt ++;            nodeList.add(node.left);            }            if(node.right != null){            cnt ++;            nodeList.add(node.right);            }            }}for(int i = 0;i < lsts.size();i++){    if(i%2 == 1){        List<Integer> ls = lsts.get(i);        int from = 0;        int to = ls.size()-1;        while(from < to){            Integer temp = ls.get(from);            ls.set(from,ls.get(to));            ls.set(to,temp);            from ++;            to --;        }    }}    return lsts;    }}

战胜了42.51，反转节点显然效率应该不是最高的。如果不采取上述的想法，我们能不能直接存入呢，当奇数层我们节点从左到右放入字节点，当为偶数层时，从右到左放入节点。

public class Solution {    public List<List<Integer>> zigzagLevelOrder(TreeNode root) {        int cnt = 0;int f = 0;int t = 1;int index; int flag = 1;int top = 0;List<List<Integer>> lsts = new ArrayList<List<Integer>>();List<Integer> lst = new ArrayList<Integer>();List<TreeNode> nodeList = new ArrayList<TreeNode>();nodeList.add(root);if(root == null){return lsts;}index = 0;while(t != 0){            if(f == t){            t = cnt;            f = 0;            cnt = 0;            lsts.add(lst);            index =top + t;            top = index;            lst = new ArrayList<Integer>();            if(flag == 1){            flag = 2;            }else{            flag = 1;            }            }            else{            TreeNode node = nodeList.get(index--);            if(node != null){            f ++;            lst.add(node.val);            }            if(flag == 1){            if(node.left != null){            cnt ++;            nodeList.add(node.left);            }            if(node.right != null){            cnt ++;            nodeList.add(node.right);            }            }else{            if(node.right != null){            cnt ++;            nodeList.add(node.right);            }            if(node.left != null){            cnt ++;            nodeList.add(node.left);            }            }            }}    return lsts;    }}

此刻战胜了92.03%

当然啦，可以递归解决的。参考大神方案

public List<List<Integer>> zigzagLevelOrder(TreeNode root)     {        List<List<Integer>> sol = new ArrayList<>();        travel(root, sol, 0);        return sol;    }        private void travel(TreeNode curr, List<List<Integer>> sol, int level)    {        if(curr == null) return;                if(sol.size() <= level)        {            List<Integer> newLevel = new LinkedList<>();            sol.add(newLevel);        }                List<Integer> collection  = sol.get(level);        if(level % 2 == 0) collection.add(curr.val);        else collection.add(0, curr.val);                travel(curr.left, sol, level + 1);        travel(curr.right, sol, level + 1);   }

效率还可以，战胜了92.09%。