問(wèn)題：
Given the root node of a binary search tree (BST) and a value to be inserted into the tree, insert the value into the BST. Return the root node of the BST after the insertion. It is guaranteed that the new value does not exist in the original BST.
Note 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.
For example,
Given the tree:
4
/ \
2 7
/ \
1 3
And the value to insert: 5
You can return this binary search tree:
4
/ \
2 7
/ \ /
1 3 5
This tree is also valid:
5
/ \
2 7
/ \
1 3
\
4
方法：
二叉搜索樹(shù)的遞歸，重點(diǎn)在于理順遞歸調(diào)用邏輯，具體實(shí)現(xiàn)參考代碼。
具體實(shí)現(xiàn)：
class InsertIntoABinarySearchTree {
// Definition for a binary tree node.
class TreeNode(var `val`: Int = 0) {
var left: TreeNode? = null
var right: TreeNode? = null
}
fun insertIntoBST(root: TreeNode?, `val`: Int): TreeNode? {
if (root == null) {
return root
}
if (`val` < root.`val`) {
if (root.left == null) {
root.left = TreeNode(`val`)
} else {
insertIntoBST(root.left, `val`)
}
} else {
if (root.right == null) {
root.right = TreeNode(`val`)
} else {
insertIntoBST(root.right, `val`)
}
}
return root
}
}
fun main(args: Array<String>) {
}
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