leetcode — recover-binary-search-tree

import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.List;

/**
 * Source : https://oj.leetcode.com/problems/recover-binary-search-tree/
 *
 *
 * 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.
 *
 * OJ's Binary Tree Serialization:
 *
 * The serialization of a binary tree follows a level order traversal, where '#' signifies
 * a path terminator where no node exists below.
 *
 * Here's an example:
 *
 *    1
 *   / \
 *  2   3
 *     /
 *    4
 *     \
 *      5
 *
 * The above binary tree is serialized as "{1,2,3,#,#,4,#,#,5}".
 */
public class RecoverBinarySearchTree {

    private TreeNode n1;
    private TreeNode n2;
    private TreeNode pre;

    /**
     * 搜索二叉树
     * 将错误调换位置的两个元素恢复位置
     *
     * 先中序遍历树，将节点的value放到一个数组中，并将节点也放到一个数组中
     * 然后将value数组排序
     * 然后依次赋值给节点数组中每个节点，然后将节点数组恢复成一棵树
     * 占用空间为O(n)
     *
     * @param root
     * @return
     */
    public TreeNode recover (TreeNode root) {
        List<Integer> arr = new ArrayList<Integer>();
        List<TreeNode> treeList = new ArrayList<TreeNode>();
        traverseInorder(root, arr, treeList);
        Collections.sort(arr);
        for (int i = 0; i < arr.size(); i++) {
            treeList.get(i).value = arr.get(i);
        }
        return root;

    }

    public void traverseInorder (TreeNode root, List<Integer> arr, List<TreeNode> treeList) {
        if (root == null) {
            return ;
        }
        traverseInorder(root.leftChild, arr, treeList);
        arr.add(root.value);
        treeList.add(root);
        traverseInorder(root.rightChild, arr, treeList);
    }

    /**
     * 二叉搜索树：中序遍历的时候是单调递增的
     *
     * 中序遍历树，将树遍历为一个链表，当前节点的值一定大于上一个节点的值，否则就是被调换的节点，中序遍历的时候记录调换的两个节点
     * 因为只有两个节点被置换，所以如果是第一次出现上一个节点的值大于当前节点，说明是被换到其前面的节点，所以被置换的是上一个节点
     * 如果是第二次出现上一个节点的值大于当前节点，那么当前节点是被置换的节点
     * 中序遍历完成后，调换记录的两个节点的值，就恢复了二叉搜索树
     *
     * @param root
     * @return
     */
    public TreeNode recoverTree (TreeNode root) {
        traverseInorder(root);
        if (n1 != null && n2 != null) {
            int temp = n1.value;
            n1.value = n2.value;
            n2.value = temp;
        }
        return root;
    }

    public void traverseInorder (TreeNode root) {
        if (root == null) {
            return;
        }
        traverseInorder(root.leftChild);
        if (pre != null) {
            if (pre.value > root.value) {
                if (n1 == null) {
                    n1 = pre;
                }
                n2 = root;
            }
        }
        pre = root;
        traverseInorder(root.rightChild);
    }

    public TreeNode createTree (char[] treeArr) {
        TreeNode[] tree = new TreeNode[treeArr.length];
        for (int i = 0; i < treeArr.length; i++) {
            if (treeArr[i] == '#') {
                tree[i] = null;
                continue;
            }
            tree[i] = new TreeNode(treeArr[i]-'0');
        }
        int pos = 0;
        for (int i = 0; i < treeArr.length && pos < treeArr.length-1; i++) {
            if (tree[i] != null) {
                tree[i].leftChild = tree[++pos];
                if (pos < treeArr.length-1) {
                    tree[i].rightChild = tree[++pos];
                }
            }
        }
        return tree[0];
    }

    /**
     * 使用广度优先遍历将树转化为数组
     *
     * @param root
     * @param chs
     */
    public void binarySearchTreeToArray (TreeNode root, List<Character> chs) {
        if (root == null) {
            chs.add('#');
            return;
        }
        List<TreeNode> list = new ArrayList<TreeNode>();
        int head = 0;
        int tail = 0;
        list.add(root);
        chs.add((char) (root.value + '0'));
        tail ++;
        TreeNode temp = null;

        while (head < tail) {
            temp = list.get(head);
            if (temp.leftChild != null) {
                list.add(temp.leftChild);
                chs.add((char) (temp.leftChild.value + '0'));
                tail ++;
            } else {
                chs.add('#');
            }
            if (temp.rightChild != null) {
                list.add(temp.rightChild);
                chs.add((char)(temp.rightChild.value + '0'));
                tail ++;
            } else {
                chs.add('#');
            }
            head ++;
        }

        //去除最后不必要的
        for (int i = chs.size()-1; i > 0; i--) {
            if (chs.get(i) != '#') {
                break;
            }
            chs.remove(i);
        }
    }

    private class TreeNode {
        TreeNode leftChild;
        TreeNode rightChild;
        int value;

        public TreeNode(int value) {
            this.value = value;
        }

        public TreeNode() {
        }
    }

    public static void main(String[] args) {
        RecoverBinarySearchTree recoverBinarySearchTree = new RecoverBinarySearchTree();
        char[] tree = new char[]{'3','4','5','#','#','2'};
        List<Character> chars = new ArrayList<Character>();
        recoverBinarySearchTree.binarySearchTreeToArray(recoverBinarySearchTree.recover(recoverBinarySearchTree.createTree(tree)), chars);
        System.out.println(Arrays.toString(chars.toArray(new Character[chars.size()])));

        chars = new ArrayList<Character>();
        recoverBinarySearchTree.binarySearchTreeToArray(recoverBinarySearchTree.recoverTree(recoverBinarySearchTree.createTree(tree)), chars);
        System.out.println(Arrays.toString(chars.toArray(new Character[chars.size()])));
    }

}