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897. 递增顺序搜索树

English Version

题目描述

给你一棵二叉搜索树的 root ，请你 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

解法

方法一：中序遍历

中序遍历过程中改变指针指向。

时间复杂度 $O(n)$。

同面试题 17.12. BiNode。

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:
        def dfs(root):
            if root is None:
                return
            nonlocal prev
            dfs(root.left)
            prev.right = root
            root.left = None
            prev = root
            dfs(root.right)

        dummy = TreeNode(val=0, right=root)
        prev = dummy
        dfs(root)
        return dummy.right

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 {
    private TreeNode prev;
    public TreeNode increasingBST(TreeNode root) {
        TreeNode dummy = new TreeNode(0, null, root);
        prev = dummy;
        dfs(root);
        return dummy.right;
    }

    private void dfs(TreeNode root) {
        if (root == null) {
            return;
        }
        dfs(root.left);
        prev.right = root;
        root.left = null;
        prev = root;
        dfs(root.right);
    }
}

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* prev;

    TreeNode* increasingBST(TreeNode* root) {
        TreeNode* dummy = new TreeNode(0, nullptr, root);
        prev = dummy;
        dfs(root);
        return dummy->right;
    }

    void dfs(TreeNode* root) {
        if (!root) return;
        dfs(root->left);
        prev->right = root;
        root->left = nullptr;
        prev = root;
        dfs(root->right);
    }
};

Go

/**
 * Definition for a binary tree node.
 * type TreeNode struct {
 *     Val int
 *     Left *TreeNode
 *     Right *TreeNode
 * }
 */
func increasingBST(root *TreeNode) *TreeNode {
	dummy := &TreeNode{Val: 0, Right: root}
	prev := dummy
	var dfs func(root *TreeNode)
	dfs = func(root *TreeNode) {
		if root == nil {
			return
		}
		dfs(root.Left)
		prev.Right = root
		root.Left = nil
		prev = root
		dfs(root.Right)
	}
	dfs(root)
	return dummy.Right
}

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