Recover Binary Search Tree

于 2015-08-23 18:29:52 发布 · 56 阅读

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本文详细介绍了如何在不改变二叉搜索树结构的情况下，解决树中两个元素位置错误的问题，提供了使用常数空间复杂度的解决方案，并解释了序列化和反序列化二叉树的概念。

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Two elements of a binary search tree (BST) are swapped by mistake.

Recover the tree without changing its structure.

Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
	private TreeNode pre;
	private TreeNode first;
	private TreeNode second;
	public void inorder(TreeNode root){
		if (root == null) {
			return;
		}
		inorder(root.left);
		if (pre == null) {
			pre = root;
		} else {
			if (pre.val >= root.val) {
				if (first == null) {
					first = pre;
				}
				second = root;
			}
			pre = root;
		}
		inorder(root.right);
	}
    public void recoverTree(TreeNode root) {
    	inorder(root);
    	if (first!=null && second!=null) {
    		int tmp = first.val;
    		first.val = second.val;
    		second.val = tmp;
    	}
    }
}