701. Insert into a Binary Search Tree - Medium
阿新 • • 發佈:2020-10-08
You are given theroot
node of a binary search tree (BST) and avalue
to insert into the tree. Returnthe root node of the BST after the insertion. It isguaranteedthat the new value does not exist in the original BST.
Noticethat there may existmultiple valid ways for theinsertion, as long as the tree remains a BST after insertion. You can returnany of them.
Example 1:
Input: root = [4,2,7,1,3], val = 5 Output: [4,2,7,1,3,5] Explanation: Another accepted tree is:
Example 2:
Input: root = [40,20,60,10,30,50,70], val = 25 Output: [40,20,60,10,30,50,70,null,null,25]
Example 3:
Input: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5 Output: [4,2,7,1,3,5]
Constraints:
- The number of nodes inthe tree will be in the range
[0,104]
. -108<= Node.val <= 108
- All the values
Node.val
areunique. -108<= val <= 108
- It'sguaranteedthat
val
does not exist in the original BST.
M1: recursion, time = O(height), space = O(height) call stack
class Solution { publicTreeNode insertIntoBST(TreeNode root, int val) { TreeNode node = new TreeNode(val); if(root == null) { return node; } if(root.val < val) { root.right = insertIntoBST(root.right, val); } else { root.left = insertIntoBST(root.left, val); } return root; } }
M2: iterative, time = O(height), space = O(1)
class Solution { public TreeNode insertIntoBST(TreeNode root, int val) { TreeNode cur = root; while(cur != null) { if(val > cur.val) { // insert into right subtree if(cur.right == null) { cur.right = new TreeNode(val); return root; } else { cur = cur.right; } } else { // insert into left subtree if(cur.left == null) { cur.left = new TreeNode(val); return root; } else { cur = cur.left; } } } return new TreeNode(val); } }