450. Delete Node in a BST.js.md

450. Delete Node in a BST

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST.

Basically, the deletion can be divided into two stages:

1. Search for a node to remove.
2. If the node is found, delete the node.

Example 1:

Input: root = [5,3,6,2,4,null,7], key = 3
Output: [5,4,6,2,null,null,7]
Explanation: Given key to delete is 3. So we find the node with value 3 and delete it.
One valid answer is [5,4,6,2,null,null,7], shown in the above BST.
Please notice that another valid answer is [5,2,6,null,4,null,7] and it's also accepted.

Example 2:

Input: root = [5,3,6,2,4,null,7], key = 0
Output: [5,3,6,2,4,null,7]
Explanation: The tree does not contain a node with value = 0.

Example 3:

Input: root = [], key = 0
Output: []

Constraints:

• The number of nodes in the tree is in the range [0, 104].
• -105 <= Node.val <= 105
• Each node has a unique value.
• root is a valid binary search tree.
• -105 <= key <= 105

方法：

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */

/**
 * @param {TreeNode} root
 * @param {number} key
 * @return {TreeNode}
 */
var deleteNode = function (root, key) {
    if (root === null) {
        return null;
    }
    if (key < root.val) {
        root.left = deleteNode(root.left, key);
    } else if (key > root.val) {
        root.right = deleteNode(root.right, key);
    } else {
        if (root.left === null) {
            const right = root.right;
            root.right = null;
            return right;
        }
        if (root.right === null) {
            const left = root.left;
            root.left = null;
            return left;
        }
        const current = findMin(root.right);
        const left = root.left;
        const right = deleteNode(root.right, current.val);
        root.left = null;
        root.right = null;
        current.left = left;
        current.right = right;
        return current;
    }
    return root;

    function findMin(root) {
        if (root.left === null) {
            return root;
        }
        return findMin(root.left);
    }
};