LintCode : Remove Node in Binary Search Tree

最新推荐文章于 2022-10-01 09:38:41 发布

weixin_30311605

最新推荐文章于 2022-10-01 09:38:41 发布

阅读量77

点赞数

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/dingjunnan/p/5417898.html

本文介绍了一种从二叉搜索树中删除指定值节点的方法。通过递归查找目标节点并考虑三种情况：节点为叶节点、仅有一个子节点、有两个子节点。对于最后一类情况，采用右子树最小值节点替代被删节点。

Description:

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.

Example:

Given binary search tree:

Remove 3, you can either return:

/**
 * Definition of TreeNode:
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left, right;
 *     public TreeNode(int val) {
 *         this.val = val;
 *         this.left = this.right = null;
 *     }
 * }
 */
public class Solution {
    /**
     * @param root: The root of the binary search tree.
     * @param value: Remove the node with given value.
     * @return: The root of the binary search tree after removal.
     */
    public TreeNode removeNode(TreeNode root, int value) {
        // write your code here
        if (root == null) {
            return null;
        }
        if (value < root.val) {
            root.left = removeNode(root.left, value);
        } else if (value > root.val) {
            root.right = removeNode(root.right, value);
        } else {
            if (root.right == null) {
                return root.left;
            }
            if (root.left == null) {
                return root.right;
            }
            TreeNode t = root;
            root = min(t.right);
            root.right = deleteMin(t.right);
            root.left = t.left;
        }
        return root;
    }
    public TreeNode deleteMin(TreeNode x) {
        if (x.left == null) {
            return x.right;
        }
        x.left = deleteMin(x.left);
        return x;
    }
    public TreeNode min(TreeNode x) {
        if (x.left == null) {
            return x;
        }
        return min(x.left);
    }
}