[leetcode]Binary Tree Level Order Traversal II

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Binary Tree Level Order Traversal II

Given a binary tree, return the bottom-up level order traversal of its nodes‘ values. (ie, from left to right, level by level from leaf to root).

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

    3   /   9  20    /     15   7

 

return its bottom-up level order traversal as:

[  [15,7],  [9,20],  [3]] 

演算法思路：

典型的BFS，與[leetcode]Binary Tree Level Order Traversal完全一樣，不過一個是對list的頭插，一個是尾插法。

代碼如下：

 1 public class Solution { 2     public List<List<Integer>> levelOrderBottom(TreeNode root) { 3         List<List<Integer>> res = new LinkedList<List<Integer>>(); 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         while(!q.isEmpty()){10             TreeNode node = q.poll();11             if(node.left != null) copy.offer(node.left);12             if(node.right != null) copy.offer(node.right);13             if(q.isEmpty() && !copy.isEmpty()){14                 List<Integer> list = new ArrayList<Integer>();15                 while(!copy.isEmpty()){16                     q.offer(copy.peek());17                     list.add(copy.poll().val);18                 }19                 res.add(0,list);20             }21         }22         return res;23     }24 }