Java for LeetCode 107 Binary Tree Level Order Traversal II

標籤：

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]]

解題思路：

修改下Java for LeetCode 102 Binary Tree Level Order Traversal即可解決，JAVA實現如下：

    public List<List<Integer>> levelOrderBottom(TreeNode root) {    List<List<Integer>> list = new ArrayList<List<Integer>>();    if (root == null)        return list;    Queue<TreeNode> queue = new LinkedList<TreeNode>();    queue.add(root);    while (queue.size() != 0) {        List<Integer> alist = new ArrayList<Integer>();        for (TreeNode child : queue)            alist.add(child.val);        list.add(new ArrayList<Integer>(alist));        Queue<TreeNode> queue2=queue;        queue=new LinkedList<TreeNode>();        for(TreeNode child:queue2){            if (child.left != null)                queue.add(child.left);            if (child.right != null)                queue.add(child.right);        }    }    Collections.reverse(list);    return list;    }

 

Java for LeetCode 107 Binary Tree Level Order Traversal II