LeetCode 题解（51）: Flatten Binary Tree to Linked List

最新推荐文章于 2025-09-08 16:25:29 发布

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最新推荐文章于 2025-09-08 16:25:29 发布

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分类专栏：算法文章标签： algorithm leetcode

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本文介绍了一种将二叉树结构通过算法转换为链表的方法，并提供了三种实现方式：C++版本使用栈进行前序遍历，Java及Python版本采用Morris算法实现空间复杂度O(1)的解决方案。

题目：

Given a binary tree, flatten it to a linked list in-place.

For example,
Given

The flattened tree should look like:

题解：自己做的笨办法，用堆栈进行前序遍历。

C++版：

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void flatten(TreeNode *root) {
        if(!root)
            return;
        if(!root->left && !root->right)
            return;
        
        stack<TreeNode*> unvisited;
        unvisited.push(root);
        while(unvisited.size()) {
            TreeNode* current = unvisited.top();
            unvisited.pop();
            if(current->right) {
                unvisited.push(current->right);
                if(current->left) {
                    unvisited.push(current->left);
                    current->right = current->left;
                    current->left = NULL;
                } 
            } else {
                if(current->left) {
                    unvisited.push(current->left);
                    current->right = current->left;
                    current->left = NULL;
                } else {
                    if(unvisited.size())
                        current->right = unvisited.top();
                    else
                        current->right = NULL;
                }
            }
        }
    }
};

网上学了一下空间复杂度O(1)算法，真正的in-place flattern。好像叫做Morris Algorithm for binary tree traversal。

Java版：

public class Solution {
    public void flatten(TreeNode root) {
        if(root == null)
            return;
        TreeNode current = root;
        while(current != null) {
            if(current.left != null) {
                TreeNode next = current.left;
                while(next.right != null)
                    next = next.right;
                next.right = current.right;
                current.right = current.left;
                current.left = null;
            }
            current = current.right;
        }
    }
}

Python版：

class Solution:
    # @param root, a tree node
    # @return nothing, do it in place
    def flatten(self, root):
        if root == None:
            return
        current = root
        while current != None:
            if current.left != None:
                next = current.left
                while next.right != None:
                    next = next.right
                next.right = current.right
                current.right = current.left
                current.left = None
            current = current.right