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Remove Node in Binary Search Tree 解答

时间:2016-10-18 11:35:16      阅读:144      评论:0      收藏:0      [点我收藏+]

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从BST中移除一个节点是比较复杂的问题,需要分好几种情况讨论。

这篇文章,就讨论了删除节点 1.有无左右子树 2.只有右子树 3.只有左子树 三种情况。

一种简单些的思维是只考虑删除节点是否有右子树(因为第一个比删除节点大的节点可能出现在右子树,不会出现在左子树)。

这里用Target表示删除节点,Parent表示删除节点的母节点。

1. 没有右子树

Parent.left / Parent.right = Target.left

2. 有右子树

1) 找到第一个比删除节点大的节点Node

2) 删除该节点Node

3) Target.val = Node.val 或 保留 Node 删除Target

/**
 * 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
        TreeNode dummy = new TreeNode(0);
        dummy.left = root;
        TreeNode parent = findNodeParent(dummy, root, value);
        TreeNode target;
        if (parent.left != null && parent.left.val == value) {
            target = parent.left;
        } else if (parent.right != null && parent.right.val == value) {
            target = parent.right;
        } else {
            return dummy.left;
        }
        deleteNode(parent, target);
        return dummy.left;
    }
    
    private TreeNode findNodeParent(TreeNode parent, TreeNode node, int val) {
        if (node == null || node.val == val) {
            return parent;
        }
        if (node.val < val) {
            return findNodeParent(node, node.right, val);
        } else {
            return findNodeParent(node, node.left, val);
        }
    }
    
    private void deleteNode(TreeNode parent, TreeNode target) {
        if (target.right == null) {
            if (parent.right == target) {
                parent.right = target.left;
            } else {
                parent.left = target.left;
            }
        } else {
            // Find first element that is greater than target
            TreeNode node1 = target.right;
            TreeNode node2 = target;
            while (node1.left != null) {
                node2 = node1;
                node1 = node1.left;
            }
            // Remove node1
            if (node1 == node2.left) {
                node2.left = node1.right;
            } else {
                node2.right = node1.right;
            }
            // Remove target
            if (target == parent.right) {
                parent.right = node1;
            } else {
                parent.left = node1;
            }
            node1.left = target.left;
            node1.right = target.right;
        }
    }
}

 

Remove Node in Binary Search Tree 解答

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原文地址:http://www.cnblogs.com/ireneyanglan/p/5972522.html

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