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].

Solution:

Mark tree by level , if it is leave then mark as level 0, then add to the List<List<int>> by level.

/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int x) { val = x; }
* }
*/
public class Solution {
public IList<IList<int>> FindLeaves(TreeNode root) {
IList<IList<int>> result = new List<IList<int>> ();
LeavesLevel(root, result);
return result;
} public int LeavesLevel(TreeNode root, IList<IList<int>> result)
{
if(root==null)
{
return -;
}
int leftLevel = LeavesLevel(root.left, result);
int rightLevel = LeavesLevel(root.right, result);
int level = Math.Max(leftLevel, rightLevel)+;
if(result.Count()<level+)
{
result.Add(new List<int>());
}
result[level].Add(root.val);
return level;
}
}

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