Binary Tree Pruning

Love this kind of tree problem: https://leetcode.com/problems/binary-tree-pruning/description/

We are given the head node root of a binary tree, where additionally every node's value is either a 0 or a 1.

Return the same tree where every subtree (of the given tree) not containing a 1 has been removed.

(Recall that the subtree of a node X is X, plus every node that is a descendant of X.)

Example 1:
Input: [1,null,0,0,1]
Output: [1,null,0,null,1]
 
Explanation: 
Only the red nodes satisfy the property "every subtree not containing a 1".
The diagram on the right represents the answer.

Example 2:
Input: [1,0,1,0,0,0,1]
Output: [1,null,1,null,1]

Example 3:
Input: [1,1,0,1,1,0,1,0]
Output: [1,1,0,1,1,null,1]

Note:

• The binary tree will have at most 100 nodes.
• The value of each node will only be 0 or 1.

Simple recursion:

• Use post-order traversal (either left-right-parent or right-left-parent)
• Use a ref variable t o indicate whether the current node needs to be pruned
• Prune the children if needed
• Before calling the recursion though, there is a special case when the tree only has the root and the root val itself is zero

That's it, thanks, ACC.

public class Solution
{
 public TreeNode PruneTree(TreeNode root)
 {
  if (root == null) return null;
  if (root.left == null && root.right == null && root.val == 0) return null;

  bool containsOne = false;
  Process(ref root, ref containsOne);
  return root;
 }

 private void Process(ref TreeNode node, ref bool containsOne)
 {
  if (node == null) return;

  bool containsOneRight = false;
  Process(ref node.right, ref containsOneRight);
  bool containsOneLeft = false;
  Process(ref node.left, ref containsOneLeft);

  containsOne = (containsOneRight || containsOneLeft || node.val == 1);
  if (!containsOneRight) node.right = null;
  if (!containsOneLeft) node.left = null;
 }
}