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

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题目：

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

For example,
Given

         1        / \       2   5      / \   \     3   4   6

The flattened tree should look like:

   1    \     2      \       3        \         4          \           5            \             6

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

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