You have a binary tree with a small defect. There is exactly one invalid node where its right child incorrectly points to another node at the same depth but to the invalid node's right.
Given the root of the binary tree with this defect, root, return the root of the binary tree after removing this invalid node and every node underneath it (minus the node it incorrectly points to).
Custom testing:
The test input is read as 3 lines:
TreeNode root
int fromNode (not available to correctBinaryTree)
int toNode (not available to correctBinaryTree)
After the binary tree rooted at root is parsed, the TreeNode with value of fromNode will have its right child pointer pointing to the TreeNode with a value of toNode. Then, root is passed to correctBinaryTree.
Example 1:
Input: root = [1,2,3], fromNode = 2, toNode = 3
Output: [1,null,3]
Explanation: The node with value 2 is invalid, so remove it.
Example 2:
Input: root = [8,3,1,7,null,9,4,2,null,null,null,5,6], fromNode = 7, toNode = 4
Output: [8,3,1,null,null,9,4,null,null,5,6]
Explanation: The node with value 7 is invalid, so remove it and the node underneath it, node 2.
Constraints:
The number of nodes in the tree is in the range [3, 104].
-109 <= Node.val <= 109
All Node.val are unique.
fromNode != toNode
fromNode and toNode will exist in the tree and will be on the same depth.
toNode is to the right of fromNode.
fromNode.right is null in the initial tree from the test data.
Solutions
Solution 1: DFS
We design a function dfs(root) to handle the subtree with root as the root. If root is null or root.right has been visited, root is an invalid node, so we return null. Otherwise, we recursively process root.right and root.left, and return root.
Finally, we return dfs(root).
The time complexity is $O(n)$, and the space complexity is $O(n)$. Here, $n$ is the number of nodes in the binary tree.
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# Definition for a binary tree node.# class TreeNode:# def __init__(self, val=0, left=None, right=None):# self.val = val# self.left = left# self.right = rightclassSolution:defcorrectBinaryTree(self,root:TreeNode)->TreeNode:defdfs(root):ifrootisNoneorroot.rightinvis:returnNonevis.add(root)root.right=dfs(root.right)root.left=dfs(root.left)returnrootvis=set()returndfs(root)