【LeetCode】897. 递增顺序查找树

题目描述：

给你一个树，请你 按中序遍历 重新排列树，使树中最左边的结点现在是树的根，并且每个结点没有左子结点，只有一个右子结点。

题目来源于Leetcode :https://leetcode-cn.com/problems/increasing-order-search-tree/

解题思路：

• 先用中序遍历将二叉树的结点值存入一个数组中，然后定义一个新的树头结点，遍历数组并将数组元素存入新树的右子树，然后新树的右子树。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    
    
    public TreeNode increasingBST(TreeNode root) {
    
    
        List<Integer> arr = new ArrayList<Integer>();
        inorder(root, arr);
        //定义新树的头结点
        TreeNode newTree = new TreeNode(0);
        TreeNode cur = newTree;
        for (int value : arr) {
    
    
            //通过遍历将原来二叉树的值存入新树的右子树
            cur.right = new TreeNode(value);
            cur = cur.right;
        }
        return newTree.right;
    }

    //中序遍历将二叉树结点值存入arr数组中
    public void inorder(TreeNode root, List<Integer> arr) {
    
    
        if (root == null) return;
        inorder(root.left, arr);
        arr.add(root.val);
        inorder(root.right, arr);
    }
}

解题思路：

• 定义一个cur当前结点，采用中序遍历的方法遍历二叉树，将二叉树的左子树结点设置为空，cur不断指向二叉树的右子树，这样一来，等二叉树遍历结束后，cur就是新的二叉树了，只要返回cur的右子树即可。

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    
    
    TreeNode cur;
    public TreeNode increasingBST(TreeNode root) {
    
    
        TreeNode newTree = new TreeNode(0);
        cur = newTree;
        inorder(root);
        return newTree.right;
    }
    public void inorder(TreeNode node) {
    
    
        if (node == null) return;
        inorder(node.left);
        node.left = null;
        cur.right = node;
        cur = node;
        inorder(node.right);
    }
}