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).
Have you met this question in a real interview?
Yes
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]
]
public class Solution {
/**
* @param root: The root of binary tree.
* @return: buttom-up level order a list of lists of integer
*/
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
// write your code here
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
if(root == null) return res;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while(!queue.isEmpty()){
int size = queue.size();
ArrayList<Integer> sub = new ArrayList<Integer>();
for(int i = 0; i < size; i++){
TreeNode node = queue.poll();
sub.add(node.val);
if(node.left != null) queue.add(node.left);
if(node.right != null) queue.add(node.right);
}
res.add(0,sub);
}
return res;
}
}
Yes
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] ]
public class Solution {
/**
* @param root: The root of binary tree.
* @return: buttom-up level order a list of lists of integer
*/
public ArrayList<ArrayList<Integer>> levelOrderBottom(TreeNode root) {
// write your code here
ArrayList<ArrayList<Integer>> res = new ArrayList<ArrayList<Integer>>();
if(root == null) return res;
Queue<TreeNode> queue = new LinkedList<TreeNode>();
queue.offer(root);
while(!queue.isEmpty()){
int size = queue.size();
ArrayList<Integer> sub = new ArrayList<Integer>();
for(int i = 0; i < size; i++){
TreeNode node = queue.poll();
sub.add(node.val);
if(node.left != null) queue.add(node.left);
if(node.right != null) queue.add(node.right);
}
res.add(0,sub);
}
return res;
}
}
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