[Lintcode]Binary Tree Zigzag Level Order Traversal

標籤：

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).

Example

Given binary tree {3,9,20,#,#,15,7},

    3   /   9  20    /     15   7

 

return its zigzag level order traversal as:

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

SOLUTION :

這題跟level order的模版都是一樣的，用queue進行BFS，只不過中間要加一些約束條件，調整level list裡面結果的順序。一般來說碰到類似奇偶迴圈的約束條件，就判斷一下(n % 2 ＝＝ 0)。很簡單的小技巧，記住就好。 

具體看代碼，備忘調整位置：

public class Solution {    /**     * @param root: The root of binary tree.     * @return: A list of lists of integer include      *          the zigzag level order traversal of its nodes‘ values      */    public ArrayList<ArrayList<Integer>> zigzagLevelOrder(TreeNode root) {        ArrayList<ArrayList<Integer>> result = new ArrayList<ArrayList<Integer>>();        if (root == null){            return result;        }        Queue<TreeNode> queue = new LinkedList<TreeNode>();        queue.offer(root);        int n = 0;// 增加進行判斷的參數        while (!queue.isEmpty()){            ArrayList<Integer> list = new ArrayList<Integer>();            int size = queue.size();            for (int i = 0; i < size; i++){                TreeNode cur = queue.remove();                //這裡進行list.add的約束                if (n % 2 == 0){                    list.add(cur.val);                } else {                    list.add(0, cur.val);                }                if (cur.left != null){                    queue.offer(cur.left);                }                if (cur.right != null){                    queue.offer(cur.right);                }            }            n++;// 以及這裡            result.add(list);        }        return result;    }}    

　　

[Lintcode]Binary Tree Zigzag Level Order Traversal