Java

  • /**
    
     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.
    
     */
    
    public class Binary_Tree_Pruning {
    
        /**
         * Definition for a binary tree node.
         * public class TreeNode {
         *     int val;
         *     TreeNode left;
         *     TreeNode right;
         *     TreeNode(int x) { val = x; }
         * }
         */
        class Solution {
            public TreeNode pruneTree(TreeNode root) {
                if (root == null) {
                    return root;
                }
    
                root.left = pruneTree(root.left);
                root.right = pruneTree(root.right);
    
                // leaf node
                if (root.left == null && root.right == null) {
                    if (root.val == 0) {
                        return null;
                    } else {
                        return root;
                    }
                }
    
                return root;
            }
        }
    }
    
  • // OJ: https://leetcode.com/problems/binary-tree-pruning/
    // Time: O(N)
    // Space: O(logN)
    class Solution {
    public:
        TreeNode* pruneTree(TreeNode* root) {
            if (!root) return NULL;
            root->left = pruneTree(root->left);
            root->right = pruneTree(root->right);
            return root->left || root->right || root->val ? root : NULL;
        }
    };
    
  • # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution(object):
        def pruneTree(self, root):
            """
            :type root: TreeNode
            :rtype: TreeNode
            """
            if not root: return
            root.left = self.pruneTree(root.left)
            root.right = self.pruneTree(root.right)
            if not root.left and not root.right and root.val == 0:
                return None
            return root
    

Java

  • /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public TreeNode pruneTree(TreeNode root) {
            if (root == null)
                return root;
            Map<TreeNode, TreeNode> childParentMap = new HashMap<TreeNode, TreeNode>();
            List<TreeNode> nodesList = new ArrayList<TreeNode>();
            Queue<TreeNode> queue = new LinkedList<TreeNode>();
            queue.offer(root);
            while (!queue.isEmpty()) {
                TreeNode node = queue.poll();
                nodesList.add(node);
                TreeNode left = node.left, right = node.right;
                if (left != null) {
                    childParentMap.put(left, node);
                    queue.offer(left);
                }
                if (right != null) {
                    childParentMap.put(right, node);
                    queue.offer(right);
                }
            }
            for (int i = nodesList.size() - 1; i >= 0; i--) {
                TreeNode node = nodesList.get(i);
                if (node.val == 0) {
                    if (node.left == null && node.right == null) {
                        TreeNode parent = childParentMap.get(node);
                        if (parent != null) {
                            if (node == parent.left)
                                parent.left = null;
                            else
                                parent.right = null;
                        } else {
                            root = null;
                            break;
                        }
                    }
                }
            }
            return root;
        }
    }
    
  • // OJ: https://leetcode.com/problems/binary-tree-pruning/
    // Time: O(N)
    // Space: O(logN)
    class Solution {
    public:
        TreeNode* pruneTree(TreeNode* root) {
            if (!root) return NULL;
            root->left = pruneTree(root->left);
            root->right = pruneTree(root->right);
            return root->left || root->right || root->val ? root : NULL;
        }
    };
    
  • # Definition for a binary tree node.
    # class TreeNode(object):
    #     def __init__(self, x):
    #         self.val = x
    #         self.left = None
    #         self.right = None
    
    class Solution(object):
        def pruneTree(self, root):
            """
            :type root: TreeNode
            :rtype: TreeNode
            """
            if not root: return
            root.left = self.pruneTree(root.left)
            root.right = self.pruneTree(root.right)
            if not root.left and not root.right and root.val == 0:
                return None
            return root
    

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