Binary Tree Level Order Traversal II LeetCode (Reverse Level Order Traversal)

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.

This problem is popular interview question in LeetCode and GeeksForGeeks A collection of hundreds of interview questions and solutions are available in our blog at Interview Question Solutions

Solution Source: Java

	import java.util.ArrayList;
	import java.util.Arrays;
	import java.util.List;
	/**
	*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,null,null,15,7],
	3
	/ \
	9 20
	/ \
	15 7
	return its bottom-up level order traversal as:
	[
	[15,7],
	[9,20],
	[3]
	]
	*
	*
	* Solution 1:
	* 1)Start from root node and for every next level, keep track of the height of the node
	* 2)Store the same height elements in a list which can be stored in another list, the height being identified by the index of the outer list
	* 3)Retrieve from the max height
	*
	* @author rbaral
	*/
	/**
	* Definition for a binary tree node.
	* public class TreeNode {
	* int val;
	* TreeNode left;
	* TreeNode right;
	* TreeNode(int x) { val = x; }
	* }
	*/
	public class BinaryOrderLevelTraversalII {
	public List<List<Integer>> levelOrderBottom(TreeNode root) {
	/*
	List result = new ArrayList();
	result.add(Arrays.asList(new int[]{5,1,15,7}));
	result.add(Arrays.asList(new int[]{9,20}));
	result.add(Arrays.asList(new int[]{3}));
	return result;
	*/
	return null;
	}
	}

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