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114. 二叉树展开为链表

题目描述

给你二叉树的根结点 root ，请你将它展开为一个单链表：

  • 展开后的单链表应该同样使用 TreeNode ，其中 right 子指针指向链表中下一个结点，而左子指针始终为 null 。
  • 展开后的单链表应该与二叉树 先序遍历 顺序相同。

 

示例 1：

输入：root = [1,2,5,3,4,null,6]
输出：[1,null,2,null,3,null,4,null,5,null,6]

示例 2：

输入：root = []
输出：[]

示例 3：

输入：root = [0]
输出：[0]

 

提示：

  • 树中结点数在范围 [0, 2000] 内
  • -100 <= Node.val <= 100

 

进阶：你可以使用原地算法（O(1) 额外空间）展开这棵树吗？

解法

方法一：寻找前驱节点

先序遍历的访问顺序是“根、左子树、右子树”，左子树最后一个节点访问完后，接着会访问根节点的右子树节点。

因此，对于当前节点，如果其左子节点不为空，我们找到左子树的最右节点，作为前驱节点，然后将当前节点的右子节点赋给前驱节点的右子节点。然后将当前节点的左子节点赋给当前节点的右子节点，并将当前节点的左子节点置为空。然后将当前节点的右子节点作为下一个节点，继续处理，直至所有节点处理完毕。

时间复杂度 $O(n)$，其中 $n$ 是树中节点的个数。空间复杂度 $O(1)$。

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# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def flatten(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        while root:
            if root.left:
                pre = root.left
                while pre.right:
                    pre = pre.right
                pre.right = root.right
                root.right = root.left
                root.left = None
            root = root.right

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/**
 * 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 void flatten(TreeNode root) {
        while (root != null) {
            if (root.left != null) {
                // 找到当前节点左子树的最右节点
                TreeNode pre = root.left;
                while (pre.right != null) {
                    pre = pre.right;
                }

                // 将左子树的最右节点指向原来的右子树
                pre.right = root.right;

                // 将当前节点指向左子树
                root.right = root.left;
                root.left = null;
            }
            root = root.right;
        }
    }
}

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    void flatten(TreeNode* root) {
        while (root) {
            if (root->left) {
                TreeNode* pre = root->left;
                while (pre->right) {
                    pre = pre->right;
                }
                pre->right = root->right;
                root->right = root->left;
                root->left = nullptr;
            }
            root = root->right;
        }
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func flatten(root *TreeNode) {
    for root != nil {
        if root.Left != nil {
            pre := root.Left
            for pre.Right != nil {
                pre = pre.Right
            }
            pre.Right = root.Right
            root.Right = root.Left
            root.Left = nil
        }
        root = root.Right
    }
}

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/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

/**
 Do not return anything, modify root in-place instead.
 */
function flatten(root: TreeNode | null): void {
    while (root !== null) {
        if (root.left !== null) {
            let pre = root.left;
            while (pre.right !== null) {
                pre = pre.right;
            }
            pre.right = root.right;
            root.right = root.left;
            root.left = null;
        }
        root = root.right;
    }
}

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// Definition for a binary tree node.
// #[derive(Debug, PartialEq, Eq)]
// pub struct TreeNode {
//   pub val: i32,
//   pub left: Option<Rc<RefCell<TreeNode>>>,
//   pub right: Option<Rc<RefCell<TreeNode>>>,
// }
//
// impl TreeNode {
//   #[inline]
//   pub fn new(val: i32) -> Self {
//     TreeNode {
//       val,
//       left: None,
//       right: None
//     }
//   }
// }
use std::rc::Rc;
use std::cell::RefCell;
impl Solution {
    #[allow(dead_code)]
    pub fn flatten(root: &mut Option<Rc<RefCell<TreeNode>>>) {
        if root.is_none() {
            return;
        }
        let mut v: Vec<Option<Rc<RefCell<TreeNode>>>> = Vec::new();
        // Initialize the vector
        Self::pre_order_traverse(&mut v, root);
        // Traverse the vector
        let n = v.len();
        for i in 0..n - 1 {
            v[i].as_ref().unwrap().borrow_mut().left = None;
            v[i].as_ref().unwrap().borrow_mut().right = v[i + 1].clone();
        }
    }

    #[allow(dead_code)]
    fn pre_order_traverse(
        v: &mut Vec<Option<Rc<RefCell<TreeNode>>>>,
        root: &Option<Rc<RefCell<TreeNode>>>
    ) {
        if root.is_none() {
            return;
        }
        v.push(root.clone());
        let left = root.as_ref().unwrap().borrow().left.clone();
        let right = root.as_ref().unwrap().borrow().right.clone();
        Self::pre_order_traverse(v, &left);
        Self::pre_order_traverse(v, &right);
    }
}

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/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {TreeNode} root
 * @return {void} Do not return anything, modify root in-place instead.
 */
var flatten = function (root) {
    while (root) {
        if (root.left) {
            let pre = root.left;
            while (pre.right) {
                pre = pre.right;
            }
            pre.right = root.right;
            root.right = root.left;
            root.left = null;
        }
        root = root.right;
    }
};

方法二

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func flatten(root *TreeNode) {
    for root != nil {
        left, right := root.Left, root.Right
        root.Left = nil
        if left != nil {
            root.Right = left
            for left.Right != nil {
                left = left.Right
            }
            left.Right = right
        }
        root = root.Right
    }
}

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