Question

701. Insert into a Binary Search Tree

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Description

You are given the root node of a binary search tree (BST) and a value to insert into the tree. Return _the root node of the BST after the insertion_. It is guaranteed that the new value does not exist in the original BST.

Notice that there may exist multiple valid ways for the insertion, as long as the tree remains a BST after insertion. You can return any 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 in the tree will be in the range [0, 104].
• -108 <= Node.val <= 108
• All the values Node.val are unique.
• -108 <= val <= 108
• It's guaranteed that val does not exist in the original BST.

Solutions

Solution 1

# 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 = right
class Solution:
  def insertIntoBST(self, root: TreeNode, val: int) -> TreeNode:
    def dfs(root):
      if root is None:
        return TreeNode(val)
      if root.val < val:
        root.right = dfs(root.right)
      else:
        root.left = dfs(root.left)
      return root

    return dfs(root)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *   int val;
 *   TreeNode left;
 *   TreeNode right;
 *   TreeNode() {}
 *   TreeNode(int val) { this.val = val; }
 *   TreeNode(int val, TreeNode left, TreeNode right) {
 *     this.val = val;
 *     this.left = left;
 *     this.right = right;
 *   }
 * }
 */
class Solution {

  public TreeNode insertIntoBST(TreeNode root, int val) {
    if (root == null) {
      return new TreeNode(val);
    }
    if (root.val < val) {
      root.right = insertIntoBST(root.right, val);
    } else {
      root.left = insertIntoBST(root.left, val);
    }
    return root;
  }
}

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *   int val;
 *   TreeNode *left;
 *   TreeNode *right;
 *   TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *   TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *   TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
  TreeNode* insertIntoBST(TreeNode* root, int val) {
    if (!root) return new TreeNode(val);
    if (root->val < val)
      root->right = insertIntoBST(root->right, val);
    else
      root->left = insertIntoBST(root->left, val);
    return root;
  }
};

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *   Val int
 *   Left *TreeNode
 *   Right *TreeNode
 * }
 */
func insertIntoBST(root *TreeNode, val int) *TreeNode {
  if root == nil {
    return &TreeNode{Val: val}
  }
  if root.Val < val {
    root.Right = insertIntoBST(root.Right, val)
  } else {
    root.Left = insertIntoBST(root.Left, val)
  }
  return root
}