Java
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/** 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; } } }
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// 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; } };
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# 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
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/** * 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; } }
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// 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; } };
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# 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