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面试题 17.12. BiNode

English Version

题目描述

二叉树数据结构TreeNode可用来表示单向链表（其中left置空，right为下一个链表节点）。实现一个方法，把二叉搜索树转换为单向链表，要求值的顺序保持不变，转换操作应是原址的，也就是在原始的二叉搜索树上直接修改。

返回转换后的单向链表的头节点。

注意：本题相对原题稍作改动

 

示例：

输入： [4,2,5,1,3,null,6,0]
输出： [0,null,1,null,2,null,3,null,4,null,5,null,6]

提示：

• 节点数量不会超过 100000。

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 897. 递增顺序查找树。

Python3

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None


class Solution:
    def convertBiNode(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(int x) { val = x; }
 * }
 */
class Solution {
    private TreeNode prev;

    public TreeNode convertBiNode(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(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* prev;

    TreeNode* convertBiNode(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 convertBiNode(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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