[leetcode]Binary Tree Zigzag Level Order Traversal

標籤：des   style   blog   http   color   io   for   ar   

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  20    /     15   7

 

return its zigzag level order traversal as:

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

演算法思路：

BFS，沒什麼好講的，用一個boolean變數標記是正序還是逆序，逆序就用頭插法，正序用尾插法。

代碼如下：

 1 public class Solution { 2      List<List<Integer>> res = new ArrayList<List<Integer>>(); 3         public List<List<Integer>> zigzagLevelOrder(TreeNode root) { 4             if(root == null) return res; 5             Queue<TreeNode> q = new LinkedList<TreeNode>(); 6             Queue<TreeNode> copy = new LinkedList<TreeNode>(); 7             q.offer(root); 8             res.add(new ArrayList<Integer>(Arrays.asList(root.val))); 9             boolean tag = false;10             while(!q.isEmpty()){11                 TreeNode tem = q.poll();12                 if(tem.left != null) copy.offer(tem.left);13                 if(tem.right != null) copy.offer(tem.right);14                 if(q.isEmpty() && !copy.isEmpty()){15                     List<Integer> list = new LinkedList<Integer>();16                     if(tag){17                         while(!copy.isEmpty()){18                             q.add(copy.peek());19                             list.add(copy.poll().val);20                         }21                     }else{22                         while(!copy.isEmpty()){23                             q.add(copy.peek());24                             list.add(0, copy.poll().val);25                         }26                     }27                     res.add(list);28                     tag = !tag;29                 }30             }31             return res;32         }33 }

 我用了兩個queue來進行層次的區分，網上很多人很巧妙的用了一個null來表示一層的結束。請參考這裡。