776. Split BST
LWC 70: 776. Split BST
Problem:
Given a Binary Search Tree (BST) with root node root, and a target value V, split the tree into two subtrees where one subtree has nodes that are all smaller or equal to the target value, while the other subtree has all nodes that are greater than the target value. It’s not necessarily the case that the tree contains a node with value V. Additionally, most of the structure of the original tree should remain. Formally, for any child C with parent P in the original tree, if they are both in the same subtree after the split, then node C should still have the parent P. You should output the root TreeNode of both subtrees after splitting, in any order.
Example 1:
Input: root = [4,2,6,1,3,5,7], V = 2 Output: [[2,1],[4,3,6,null,null,5,7]] Explanation: Note that root, output[0], and output1 are TreeNode objects, not arrays. The given tree [4,2,6,1,3,5,7] is represented by the following diagram:
4
/ \
2 6
/ \ / \
1 3 5 7while the diagrams for the outputs are:
4
/ \
3 6 and 2
/ \ /
5 7 1Note:
- The size of the BST will not exceed 50.
- The BST is always valid and each node’s value is different.
思路： 问题的关键在于进行切分后，树的结构不能改变。影响BST的结构在于输入顺序，所以切分前后，维持输入顺序即可。而BST的层序遍历刚好是最初的输入顺序。所以 1. 求出输入顺序。 2. 根据val划分两个子输入顺序 3. 根据子输入顺序建立BST
代码如下：
public TreeNode[] splitBST(TreeNode root, int V) {
if (root == null) return new TreeNode[] {null, null};
List<Integer> nodes = new ArrayList<>();
Queue<TreeNode> q = new ArrayDeque<>();
q.offer(root);
while (!q.isEmpty()) {
TreeNode now = q.poll();
nodes.add(now.val);
if (now.left != null) {
q.offer(now.left);
}
if (now.right != null) {
q.offer(now.right);
}
}
List<Integer> left = new ArrayList<>();
List<Integer> right = new ArrayList<>();
for (int val : nodes) {
if (val <= V) left.add(val);
else right.add(val);
}
TreeNode leftNode = null;
for (int val : left) leftNode = build(leftNode, val);
TreeNode rightNode = null;
for (int val : right) rightNode = build(rightNode, val);
return new TreeNode[] {leftNode, rightNode};
}
TreeNode build(TreeNode root, int val) {
if (root == null) return new TreeNode(val);
else {
if (val <= root.val) {
root.left = build(root.left, val);
}
else {
root.right = build(root.right, val);
}
return root;
}
}再来一种递归的思路，很简单。我们把问题进行拆分合并式求解。比如当root.val <= val时，说明root和root的左子树均小于等于val，这种结构维持，且以root为根，那么问题就转而求root右子树的分裂。因为分裂的第一颗树，是小于val的，所以需要链接到root的右子树上，详见代码。
Java版本：
public TreeNode[] splitBST(TreeNode root, int V) {
if (root == null) return new TreeNode[] {null, null};
if (root.val <= V) {
TreeNode[] res = splitBST(root.right, V);
root.right = res[0];
res[0] = root;
return res;
}
else{
TreeNode[] res = splitBST(root.left, V);
root.left = res[1];
res[1] = root;
return res;
}
}Python版本：
# 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 splitBST(self, root, V):
"""
:type root: TreeNode
:type V: int
:rtype: List[TreeNode]
"""
if not root: return [None] * 2
if root.val <= V:
res = self.splitBST(root.right, V)
root.right = res[0]
res[0] = root
return res
else:
res = self.splitBST(root.left, V)
root.left = res[1]
res[1] = root
return res