README.md

814. Binary Tree Pruning

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.

Solutions (Python)

1. Recursion

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def pruneTree(self, root: TreeNode) -> TreeNode:
        if not root:
            return None

        root.left = self.pruneTree(root.left)
        root.right = self.pruneTree(root.right)

        if root.val == 1 or root.left or root.right:
            return root
        else:
            return None

Solutions (Ruby)

1. Recursion

# Definition for a binary tree node.
# class TreeNode
#     attr_accessor :val, :left, :right
#     def initialize(val = 0, left = nil, right = nil)
#         @val = val
#         @left = left
#         @right = right
#     end
# end
# @param {TreeNode} root
# @return {TreeNode}
def prune_tree(root)
    return nil if root.nil?

    root.left = prune_tree(root.left)
    root.right = prune_tree(root.right)

    if root.val == 1 or not root.left.nil? or not root.right.nil?
        return root
    else
        return nil
    end
end