树
二叉搜索树
二叉树
题目描述
给定二叉搜索树(BST)的根节点 root
和要插入树中的值 value
,将值插入二叉搜索树。 返回插入后二叉搜索树的根节点。 输入数据 保证 ,新值和原始二叉搜索树中的任意节点值都不同。
注意 ,可能存在多种有效的插入方式,只要树在插入后仍保持为二叉搜索树即可。 你可以返回 任意有效的结果 。
示例 1:
输入: root = [4,2,7,1,3], val = 5
输出: [4,2,7,1,3,5]
解释: 另一个满足题目要求可以通过的树是:
示例 2:
输入: root = [40,20,60,10,30,50,70], val = 25
输出: [40,20,60,10,30,50,70,null,null,25]
示例 3:
输入: root = [4,2,7,1,3,null,null,null,null,null,null], val = 5
输出: [4,2,7,1,3,5]
提示:
树中的节点数将在 [0, 104 ]
的范围内。
-108 <= Node.val <= 108
所有值 Node.val
是 独一无二 的。
-108 <= val <= 108
保证 val
在原始BST中不存在。
解法
方法一
Python3 Java C++ Go TypeScript
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18 # 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 )
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29 /**
* 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 ;
}
}
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22 /**
* 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 ;
}
};
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19 /**
* 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
}
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22 /**
* Definition for a binary tree node.
* class TreeNode {
* val: number
* left: TreeNode | null
* right: TreeNode | null
* constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
* this.val = (val===undefined ? 0 : val)
* this.left = (left===undefined ? null : left)
* this.right = (right===undefined ? null : right)
* }
* }
*/
function insertIntoBST ( root : TreeNode | null , val : number ) : TreeNode | null {
if ( ! root ) return new TreeNode ( val );
if ( val < root . val ) root . left = insertIntoBST ( root . left , val );
else root . right = insertIntoBST ( root . right , val );
return root ;
}
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