[leetcode]Binary Tree Level Order Traversal II

问题描述：

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]
]

confused what

"{1,#,2,3}"

means?> read more on how binary tree is serialized
on OJ.

代码：

public List<List<Integer>> levelOrderBottom(TreeNode root) {  //java
if(root == null)
return new ArrayList<>();

Stack<TreeNode> stack = new Stack<TreeNode>();
stack.add(root);
List<TreeNode> levelList = new ArrayList<TreeNode>();

List<List<Integer>> result = new ArrayList<List<Integer>>();
List<List<Integer>> invert_result = new ArrayList<List<Integer>>();
List<Integer> tmp = new ArrayList<Integer>();

while(!stack.isEmpty()|| !levelList.isEmpty()){
if(stack.isEmpty()){
for(int i=levelList.size()-1; i>=0; i--)
stack.push(levelList.get(i));
levelList.clear();
invert_result.add(tmp);
tmp = new ArrayList<Integer>();
}

while(!stack.isEmpty()){
TreeNode node = stack.pop();
tmp.add(node.val);

if(node.left !=null)
levelList.add(node.left);
if(node.right !=null)
levelList.add(node.right);
}
}
invert_result.add(tmp);

//invert result
for(int i=invert_result.size()-1; i>=0; i--)
result.add(invert_result.get(i));

return result;

}