Remove Node in Binary Search Tre

Remove Node in Binary Search Tree

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今天是一道题目，来自LintCode，难度为Hard，Acceptance为25%。

题目如下

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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:

          5
       /    \
    3          6
 /    \
2       4

Remove 3, you can either return:

          5
       /    \
    2          6
      \
         4

or :

          5
       /    \
    4          6
 /   
2

解题思路及代码见阅读原文

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解题思路

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首先，该题在《算法导论》一书中是有的。思路也较为容易理解，关键是实现算法时需要注意的各种细节。

那么下面说明该题的思路：

第一步，当然是找到要删除的节点node和它的父节点parent，这里采用前序遍历即可。如题目中所说，如果找到了就继续，否则什么都不做，直接返回。

第二步，就是删除该node节点，这里分为两种情况：

• node节点的右子树不存在。此时，直接用parent节点的子节点指向该node节点的左节点。

• node节点的右子树存在。此时，需要找到右子树的最小节点，用该节点替换node节点。

技巧：
因为要删除的节点可能是根节点，因此为了算法的通用性，可以首先new一个dummy节点，该节点的左节点指向根节点，这样处理起来更为方便。

最后，我们来看代码。

代码如下

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Java版

/**
 * 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) {
        TreeNode dummy = new TreeNode(0);
        dummy.left = root;
        
        TreeNode parent = findNode(dummy, root, value);
        TreeNode node;
        if (parent.left != null && parent.left.val == value) {
            node = parent.left;
        } else if (parent.right != null && parent.right.val == value) {
            node = parent.right;
        } else {
            return dummy.left;
        }
        
        deleteNode(parent, node);
        
        return dummy.left;
    }
    
    private TreeNode findNode(TreeNode parent, TreeNode node, int value) {
        if (node == null) {
            return parent;
        }
        
        if (node.val == value) {
            return parent;
        }
        if (value < node.val) {
            return findNode(node, node.left, value);
        } else {
            return findNode(node, node.right, value);
        }
    }
    
    private void deleteNode(TreeNode parent, TreeNode node) {
        if (node.right == null) {
            if (parent.left == node) {
                parent.left = node.left;
            } else {
                parent.right = node.left;
            }
        } else {
            TreeNode temp = node.right;
            TreeNode father = node;
            
            while (temp.left != null) {
                father = temp;
                temp = temp.left;
            }
            
            if (father.left == temp) {
                father.left = temp.right;
            } else {
                father.right = temp.right;
            }
            
            if (parent.left == node) {
                parent.left = temp;
            } else {
                parent.right = temp;
            }
            
            temp.left = node.left;
            temp.right = node.right;
        }
    }
}