450. 删除二叉搜索树中的节点

给定一个二叉搜索树的根节点 root 和一个值 key，删除二叉搜索树中的 key 对应的节点，并保证二叉搜索树的性质不变。返回二叉搜索树（有可能被更新）的根节点的引用。

一般来说，删除节点可分为两个步骤：

首先找到需要删除的节点；
如果找到了，删除它。
 
来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/delete-node-in-a-bst
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

递归

class Solution {

    private TreeNode successor(TreeNode root) {
        TreeNode node = root.right;
        while (node.left != null) {
            node = node.left;
        }
        return node;
    }

    private TreeNode predecessor(TreeNode root) {
        TreeNode node = root.left;
        while (node.right != null) {
            node = node.right;
        }
        return node;
    }

    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) {
            return null;
        }
        if (root.val < key) {
            root.right = deleteNode(root.right, key);
        } else if (root.val > key) {
            root.left = deleteNode(root.left, key);
        } else {
            if (root.left == null && root.right == null) {
                root = null;
            } else if (root.right != null) {
                root.val = successor(root).val;
                root.right = deleteNode(root.right, root.val);
            } else {
                root.val = predecessor(root).val;
                root.left = deleteNode(root.left, root.val);
            }
        }
        return root;
    }
}


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;
    }
}

迭代

class Solution {
    public TreeNode deleteNode(TreeNode root, int key) {
        if (root == null) {
            return null;
        }
        TreeNode cur = root, curParent = null;
        while (cur != null && cur.val != key) {
            curParent = cur;
            if (cur.val > key) {
                cur = cur.left;
            } else {
                cur = cur.right;
            }
        }
        if (cur == null) {
            return root;
        }
        if (cur.left == null && cur.right == null) {
            cur = null;
        } else if (cur.left == null) {
            cur = cur.right;
        } else if (cur.right == null) {
            cur = cur.left;
        } else {
            TreeNode precessor = cur.left, precessorParent = cur;
            while (precessor.right != null) {
                precessorParent = precessor;
                precessor = precessor.right;
            }
            if (precessorParent.val == cur.val) {
                precessorParent.left = precessor.left;
            } else {
                precessorParent.right = precessor.left;
            }
            precessor.left = cur.left;
            precessor.right = cur.right;
            cur = precessor;
        }

        if (curParent == null) {
            return cur;
        } else {
            if (curParent.left != null && curParent.left.val == key) {
                curParent.left = cur;
            } else {
                curParent.right = cur;
            }
        }
        return root;
    }
}