【Leetcode】Flatten Binary Tree to Linked List

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

/**
 * 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 == nullptr) {
            return;
        }
        if (root->left != nullptr) {
            flatten(root->left);
        }
        if (root->right != nullptr) {
            flatten(root->right);
        }
        if (root->left != nullptr) {
            TreeNode *tail = root->left;
            while (tail->right != nullptr) {
                tail = tail->right;
            }
            tail->right = root->right;
            root->right = root->left;
            root->left = nullptr;
        }
    }
};

flatten后的链表顺序与树的先序遍历一致。树的问题通常可以考虑递归。

也可以借助栈迭代求解：

/**
 * 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 == nullptr) {
            return;
        }
        stack<TreeNode *> s;
        s.push(root);
        while (!s.empty()) {
            TreeNode *p = s.top();
            s.pop();
            if (p->right != nullptr) {
                s.push(p->right);
            }
            if (p->left != nullptr) {
                s.push(p->left);
            }
            if (!s.empty()) {
                p->right = s.top();
            }
            p->left = nullptr;
        }
    }
};