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Binary Tree Level Order Traversal II.java
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Binary Tree Level Order Traversal II.java
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/*
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).
Link: https://leetcode.com/problems/binary-tree-level-order-traversal-ii/
Example: 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]
]
Solution: None
Source: None
*/
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
List<List<Integer>> result = new ArrayList<List<Integer>>();
if(root == null){
return result;
}
LinkedList<TreeNode> current = new LinkedList<TreeNode>();
LinkedList<TreeNode> next = new LinkedList<TreeNode>();
current.offer(root);
ArrayList<Integer> numberList = new ArrayList<Integer>();
// need to track when each level starts
while(!current.isEmpty()){
TreeNode head = current.poll();
numberList.add(head.val);
if(head.left != null){
next.offer(head.left);
}
if(head.right!= null){
next.offer(head.right);
}
if(current.isEmpty()){
current = next;
next = new LinkedList<TreeNode>();
result.add(numberList);
numberList = new ArrayList<Integer>();
}
}
//return Collections.reverse(result);
List<List<Integer>> reversedResult = new ArrayList<List<Integer>>();
for(int i = result.size() - 1; i >= 0; i--){
reversedResult.add(result.get(i));
}
return reversedResult;
}
}