Day-2 Binary Tree Level Order Traversal II

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> result=new ArrayList<>();
        if(root==null) return result;
        Queue<TreeNode> st=new LinkedList<>();
        
        st.add(root);
        int count=1;
        while(!st.isEmpty()){
            
            List<Integer> l=new ArrayList<>();
            int num=0;
            
            while(count-->0){
                TreeNode t=st.remove();
                if(t.left!=null){
                    st.add(t.left);
                    num++;
                }
                if(t.right!=null){
                    st.add(t.right);
                    num++;
                }
                l.add(t.val);
            }
            
            count=num;
            result.add(0,l);
        }
        
        return result;
    }
}

class Solution {
    public List<List<Integer>> levelOrderBottom(TreeNode root) {
        List<List<Integer>> result=new ArrayList<>();
        if(root==null) return result;
        
        dfs(root,0,result);
        return result;
    }
    
    void dfs(TreeNode root, int depth,List<List<Integer>> result){
        if(root==null) return;
        if(result.size()==depth){
            result.add(0,new ArrayList<>());
        }
        
        result.get(result.size()-depth-1).add(root.val);
        dfs(root.left,depth+1,result);
        dfs(root.right,depth+1,result);
        
    }
}