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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.com/problems/increasing-order-search-tree/

解法

由于二叉搜索树的性质，可以利用中序遍历得到递增序列

Python3

# 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

Java

/**
 * 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;
    }
}

Go

/**
 * 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
}

C++

/**
 * 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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