366. Find Leaves of Binary Tree

Given a binary tree, collect a tree's nodes as if you were doing this: Collect and remove all leaves, repeat until the tree is empty.

Example:
Given binary tree 

          1
         / 
        2   3
       /      
      4   5    

Returns [4, 5, 3], [2], [1].

Explanation:

1. Removing the leaves [4, 5, 3] would result in this tree:

          1
         / 
        2          

2. Now removing the leaf [2] would result in this tree:

          1          

3. Now removing the leaf [1] would result in the empty tree:

          []         

Returns [4, 5, 3], [2], [1].

 解法1 的思路是用DFS的方法，求出最大深度，每次都要在结果上加上相对应的值。

public IList<IList<int>> FindLeaves(TreeNode root) {
        var res = new List<IList<int>>();
        
        if(root == null) return res;
       DFS(root,res);
       
        return res;
    }
    
    public int DFS(TreeNode root,IList<IList<int>> res )
    {
        if(root == null) return -1;
        int a = 1+Math.Max(DFS(root.left,res), DFS(root.right,res));
        if(a >= res.Count()) res.Add(new List<int>());
        res[a].Add(root.val);
        return a;
    }

 另一种解法是删除leaves。一直到root

public IList<IList<int>> FindLeaves(TreeNode root) {
        var res = new List<IList<int>>();
        if(root == null) return res;
        while(root != null)
        {
            var cur = new List<int>(); 
            root = DFS(root,cur);
            res.Add(cur);
        }
        return res;
    }
    
    public TreeNode DFS(TreeNode root,IList<int> res )
    {
        if(root == null) return null;
        if(root.left == null && root.right == null)
        {
            res.Add(root.val);
            return null;
        }
        root.left = DFS(root.left, res);
        root.right = DFS(root.right,res);
        return root;
    }