LeetCode | Binary Tree Level Order Traversal II

题目

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],
]

分析
在Binary Tree Level Order Traversal基础上稍作修改。

代码

import java.util.ArrayList;
import java.util.LinkedList;
import java.util.Queue;

public class BinaryTreeLevelOrderTraversalII {
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
LinkedList<ArrayList<Integer>> results = new LinkedList<ArrayList<Integer>>();
Queue<TreeNode> curLevel = new LinkedList<TreeNode>();
Queue<TreeNode> nextLevel = new LinkedList<TreeNode>();
if (root == null) {
return new ArrayList<ArrayList<Integer>>();
}
curLevel.add(root);
ArrayList<Integer> list = new ArrayList<Integer>();
while (!curLevel.isEmpty()) {
TreeNode node = curLevel.poll();
list.add(node.val);
if (node.left != null) {
nextLevel.add(node.left);
}
if (node.right != null) {
nextLevel.add(node.right);
}
if (curLevel.isEmpty()) {
Queue<TreeNode> temp = curLevel;
curLevel = nextLevel;
nextLevel = temp;
results.addFirst(list);
list = new ArrayList<Integer>();
}
}

return new ArrayList<ArrayList<Integer>>(results);
}
}