Java for LeetCode 107 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]
]

解题思路:

修改下Java for LeetCode 102 Binary Tree Level Order Traversal即可解决,JAVA实现如下:

    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;
}
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