Question
Formatted question description: https://leetcode.ca/all/107.html
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]
]
@tag-tree
Algorithm
The traversal from the bottom sequence is actually traversal from the top, but the final storage method has changed.
Code
Java
import java.util.*;
public class Binary_Tree_Level_Order_Traversal_II {
/**
* 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>> list = new ArrayList<List<Integer>>();
if (root == null) {
return list;
}
Queue<TreeNode> sk = new LinkedList<>();
sk.offer(root);
sk.offer(null);// @note: use null as marker for end of level
List<Integer> oneLevel = new ArrayList<>();
while (!sk.isEmpty()) {
TreeNode current = sk.poll();
if (current == null) {
List<Integer> copy = new ArrayList<>(oneLevel);
list.add(0, copy); // diff of I and II: insert to head
// clean after one level recorded
oneLevel.clear();// @memorize: this api
// @note:@memorize: if stack is now empty then DO NOT add null, or else infinite looping
// sk.offer(null); // add marker
if (!sk.isEmpty()) {
sk.offer(null); // add marker
}
continue;
}
oneLevel.add(current.val);
// @note:@memorize: since using null as marker, then must avoid adding null when children are null
// sk.offer(current.left);
// sk.offer(current.right);
if (current.left != null) {
sk.offer(current.left);
}
if (current.right != null) {
sk.offer(current.right);
}
}
return list;
}
}
public class Solution_dummyNodeAsMarker {
public List<List<Integer>> levelOrderBottom(TreeNode root) {
LinkedList<List<Integer>> result = new LinkedList<List<Integer>>();
if(root == null) {
return result;
}
// Array deques have no capacity restrictions, they grow as necessary
Queue<TreeNode> queue = new ArrayDeque<TreeNode>(); // @note: another implementation
queue.add(root);
TreeNode dummy = new TreeNode(0);
queue.add(dummy);
List<Integer> currentLevel = new ArrayList<Integer>();
while(queue.size() > 1){
TreeNode current = queue.poll();
if (current.hashCode() != dummy.hashCode()) { // more specific with hashCode()
currentLevel.add(current.val);
if(current.left != null) queue.add(current.left);
if(current.right != null) queue.add(current.right);
} else {
result.addFirst(new ArrayList<Integer>(currentLevel));
currentLevel.clear();
queue.add(dummy);
}
}
// @note: LinkedList has API addFirst()
result.addFirst(currentLevel);
// or call reverse()
// Collections.reverse(result);
return result;
}
}
}