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] ]
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public IList<IList<int>> LevelOrderBottom(TreeNode root) {
var result = new List<IList<int>>();
var queue = new Queue<TreeNode>();
var stack = new Stack<IList<int>>();
if(root == null){
return result;
}
queue.Enqueue(root);
while(queue.Count()>0){
int count = queue.Count();
var innerRes = new List<int>();
while(count-->0){
var temp = queue.Dequeue();
innerRes.Add(temp.val);
if(temp.left != null){
queue.Enqueue(temp.left);
}
if(temp.right != null){
queue.Enqueue(temp.right);
}
}
stack.Push(innerRes);
}
while(stack.Count()>0){
result.Add(stack.Pop());
}
return result;
}
}Time Complexity: O(n) in the worst case
Space Complexity: O(1)
