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剑指 Offer II 052. 展平二叉搜索树

题目描述

给你一棵二叉搜索树，请 按中序遍历 将其重新排列为一棵递增顺序搜索树，使树中最左边的节点成为树的根节点，并且每个节点没有左子节点，只有一个右子节点。

 

示例 1：

输入：root = [5,3,6,2,4,null,8,1,null,null,null,7,9]
输出：[1,null,2,null,3,null,4,null,5,null,6,null,7,null,8,null,9]

示例 2：

输入：root = [5,1,7]
输出：[1,null,5,null,7]

 

提示：

  • 树中节点数的取值范围是 [1, 100]
  • 0 <= Node.val <= 1000

 

注意：本题与主站 897 题相同： https://leetcode.cn/problems/increasing-order-search-tree/

解法

方法一

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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 increasingBST(self, root: TreeNode) -> TreeNode:
        head, tail = None, None
        stack = []
        cur = root
        while stack or cur:
            while cur:
                stack.append(cur)
                cur = cur.left
            cur = stack.pop()
            if not head:
                head = cur
            else:
                tail.right = cur
            tail = cur
            cur.left = None
            cur = cur.right
        return head

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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 TreeNode increasingBST(TreeNode root) {
        TreeNode head = null, tail = null;
        Deque<TreeNode> stack = new ArrayDeque<>();
        TreeNode cur = root;
        while (!stack.isEmpty() || cur != null) {
            while (cur != null) {
                stack.push(cur);
                cur = cur.left;
            }
            cur = stack.pop();
            if (head == null) {
                head = cur;
            } else {
                tail.right = cur;
            }
            tail = cur;
            cur.left = null;
            cur = cur.right;
        }
        return head;
    }
}

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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:
    TreeNode* increasingBST(TreeNode* root) {
        TreeNode *head = nullptr, *tail = nullptr;
        stack<TreeNode*> stk;
        TreeNode* cur = root;
        while (!stk.empty() || cur != nullptr) {
            while (cur != nullptr) {
                stk.push(cur);
                cur = cur->left;
            }
            cur = stk.top();
            stk.pop();
            if (head == nullptr) {
                head = cur;
            } else {
                tail->right = cur;
            }
            tail = cur;
            cur->left = nullptr;
            cur = cur->right;
        }
        return head;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func increasingBST(root *TreeNode) *TreeNode {
    var head, tail *TreeNode
    stack := make([]*TreeNode, 0)
    cur := root
    for len(stack) > 0 || cur != nil {
        for cur != nil {
            stack = append(stack, cur)
            cur = cur.Left
        }
        cur = stack[len(stack)-1]
        stack = stack[:len(stack)-1]
        if head == nil {
            head = cur
        } else {
            tail.Right = cur
        }
        tail = cur
        cur.Left = nil
        cur = cur.Right
    }
    return head
}

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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)
 *     }
 * }
 */

function increasingBST(root: TreeNode | null): TreeNode | null {
    const dummy = new TreeNode();
    let cur = dummy;
    const dfs = (root: TreeNode | null) => {
        if (root == null) {
            return;
        }
        dfs(root.left);
        cur.right = new TreeNode(root.val);
        cur = cur.right;
        dfs(root.right);
    };
    dfs(root);
    return dummy.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 {
    fn dfs(root: &Option<Rc<RefCell<TreeNode>>>, vals: &mut Vec<i32>) {
        if root.is_none() {
            return;
        }
        let node = root.as_ref().unwrap().borrow();
        Self::dfs(&node.left, vals);
        vals.push(node.val);
        Self::dfs(&node.right, vals);
    }

    pub fn increasing_bst(root: Option<Rc<RefCell<TreeNode>>>) -> Option<Rc<RefCell<TreeNode>>> {
        let mut vals = Vec::new();
        Self::dfs(&root, &mut vals);
        let mut dummy = Rc::new(RefCell::new(TreeNode::new(0)));
        for &val in vals.iter().rev() {
            let mut dummy = dummy.as_ref().borrow_mut();
            dummy.right = Some(
                Rc::new(
                    RefCell::new(TreeNode {
                        val,
                        left: None,
                        right: dummy.right.take(),
                    })
                )
            );
        }
        let ans = dummy.as_ref().borrow_mut().right.take();
        ans
    }
}

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/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     struct TreeNode *left;
 *     struct TreeNode *right;
 * };
 */

struct TreeNode* dfs(struct TreeNode* root, struct TreeNode* cur) {
    if (!root) {
        return cur;
    }
    cur = dfs(root->left, cur);
    cur->right = malloc(sizeof(struct TreeNode));
    cur->right->val = root->val;
    cur->right->left = NULL;
    cur->right->right = NULL;
    cur = cur->right;
    return dfs(root->right, cur);
}

struct TreeNode* increasingBST(struct TreeNode* root) {
    struct TreeNode* dummy = malloc(sizeof(struct TreeNode));
    dfs(root, dummy);
    return dummy->right;
}

方法二

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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 increasingBST(self, root: TreeNode) -> TreeNode:
        def dfs(root: TreeNode):
            if root is None:
                return

            dfs(root.left)

            nonlocal cur
            cur.right = root
            root.left = None
            cur = cur.right

            dfs(root.right)

        cur = dummy = TreeNode()
        dfs(root)
        return dummy.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 {
    private TreeNode cur;

    public TreeNode increasingBST(TreeNode root) {
        TreeNode dummy = new TreeNode();
        cur = dummy;
        dfs(root);
        return dummy.right;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        cur.right = root;
        root.left = null;
        cur = cur.right;
        dfs(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:
    TreeNode* increasingBST(TreeNode* root) {
        TreeNode* dummy = new TreeNode();
        TreeNode* cur = dummy;
        function<void(TreeNode*)> dfs = [&](TreeNode* root) {
            if (!root) {
                return;
            }
            dfs(root->left);
            cur->right = root;
            root->left = nullptr;
            cur = cur->right;
            dfs(root->right);
        };
        dfs(root);
        return dummy->right;
    }
};

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/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func increasingBST(root *TreeNode) *TreeNode {
    dummy := &TreeNode{}
    cur := dummy
    var dfs func(*TreeNode)
    dfs = func(root *TreeNode) {
        if root == nil {
            return
        }
        dfs(root.Left)
        root.Left = nil
        cur.Right = root
        cur = root
        dfs(root.Right)
    }
    dfs(root)
    return dummy.Right
}

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