【C++】【LeetCode】114. Flatten Binary Tree to Linked List

题目

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

思路

有两种方法，第一种比较笨，把所有的点按照前序遍历存入队列，然后取出来构建一棵树；还有一种，把左子树的最右叶子节点找到，将右子树作为最右叶子节点的右子树，然后将左子树作为右子树。

代码

思路一/** * 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:    void flatten(TreeNode* root) {        if (root == NULL) {            return ;        }        std::queue<TreeNode*> s;        helper(s, root);        TreeNode *head = root;        s.pop();        while (s.size() > 0) {            head->right = s.front();            head->left = NULL;            head = head->right;            s.pop();        }    }    void helper(std::queue<TreeNode*>& s, TreeNode* root) {        if (root == NULL) {            return;        }        s.push(root);        helper(s, root->left);        helper(s, root->right);    }};

思路二class Solution {public:    void flatten(TreeNode *root) {        TreeNode*now = root;        while (now)        {            if(now->left)            {                //Find current node's prenode that links to current node's right subtree                TreeNode* pre = now->left;                while(pre->right)                {                    pre = pre->right;                }                pre->right = now->right;                //Use current node's left subtree to replace its right subtree(original right                 //subtree is already linked by current node's prenode                now->right = now->left;                now->left = NULL;            }            now = now->right;        }    }};