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Recover Binary Search Tree@LeetCode

LeetCode
Word count: 257 / Reading time: 1 min
2015/04/19 Share

Recover Binary Search Tree

根据BST树的特性来，对BST的中序遍历，得到的是一个升序数列。所以在遍历过程中检测出两个异常的位置，对其进行交换即可。

一旦有两个位置的节点被交换了，那么中序遍历就会出现有两个：Node[i] > Node[i + 1]其中i是错误位置，Node[j] < Node[j - 1]其中j是错误位置，遵循这个规律，找到相应的Node[i]和Node[j]对其进行交换（只交换val值）即可。

实现代码如下：

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public class Solution {

    private TreeNode wrongLessNode;
    private TreeNode wrongLargerNode;
    private TreeNode preNode;

    public void recoverTree(TreeNode root) {
        recover(root);
        if (wrongLessNode != null && wrongLargerNode != null) {
            int temp = wrongLessNode.val;
            wrongLessNode.val = wrongLargerNode.val;
            wrongLargerNode.val = temp;
        }
    }

    private void recover(TreeNode root) {
        if (root == null)
            return;
        if (preNode == null && root.left == null) {
            preNode = root;
        }
        recover(root.left);
        if (preNode != null && root.val < preNode.val) {
            if (wrongLessNode == null) {
                wrongLessNode = preNode;
                wrongLargerNode = root;
            }
            else {
                wrongLargerNode = root;
                return;
            }
        }
        preNode = root;
        recover(root.right);
    }
}

原文作者: findingsea

原文链接: http://findingsea.github.io/2015/04/19/recover-binary-search-tree-at-leetcode/

发表日期: April 19th 2015, 9:47:00 am

版权声明: 本文采用知识共享署名-非商业性使用 4.0 国际许可协议进行许可

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