450.Delete-Node-in-a-BST

450. Delete Node in a BST

题目地址

https://leetcode.com/problems/delete-node-in-a-bst/

http://www.lintcode.com/problem/remove-node-in-binary-search-tree/

https://www.jiuzhang.com/solution/remove-node-in-binary-search-tree/

http://www.mathcs.emory.edu/~cheung/Courses/171/Syllabus/9-BinTree/BST-delete.html

http://www.mathcs.emory.edu/~cheung/Courses/171/Syllabus/9-BinTree/BST-delete2.html

题目描述

Given a root of Binary Search Tree with unique value for each node. Remove the node with given value. If there is no such a node with given value in the binary search tree, do nothing. You should keep the tree still a binary search tree after removal.

代码

Approach 1: Recursion

public class Solution {
  public int successor(TreeNode root) {
    root = root.right;
    while (root.left != null) root = root.left;
    return root.val;
  }

  public int predecessor(TreeNode root) {
    root = root.left;
    while (root.right != null) root = root.right;
    return root.val;
  }

  public TreeNode deleteNote(TreeNode root, int key) {
    if (root == null) return null;

    // delete from the right subtree
    if (key > root.val) {
      root.right = deleteNode(root.right, key);
    } else if (ke < root.val) {
      root.left = deleteNode(root.left, key);
    } else {
            // root.val == key
      if (root.left == null && root.right == null) {
        // the node is a leaf
        root = null;
      } else if (root.right != null){
        // the node is not a leaf and has a right child
        root.val = successor(root);
        root.right = deleteNode(root.right, root.val);
      } else {
        // the node is not a leaf, has no right child, and has left child
        root.val = predecessor(root);
        root.left = deleteNode(root.left, root.val);
      }
    }

    return root;
  }
}

Time complexity : $O(logN)$

Space complexity : $O(logN)$