Leetcode Flatten Binary Tree to Linked List

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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

此題看題目給出的樣本就知道是先序遍曆
本題用遞迴求先序遍曆

class Solution {public:    vector<TreeNode *> res;        void preTraverse(TreeNode *root){        if(root == NULL) return;        res.push_back(root);        preTraverse(root->left);        preTraverse(root->right);    }        void flatten(TreeNode *root) {        if(root == NULL) return;        preTraverse(root);        for(int i = 1 ; i < res.size(); ++ i){            res[i-1]->left =NULL;            res[i-1]->right = res[i];        }        res[res.size()-1]->left =NULL;        res[res.size()-1]->right = NULL;    }};

遞迴求解

下面用迭代求解，一步步的構造結點，由於題目是先序遍曆，可以將右子樹連結到左子樹最右的孩子結點，如果不明白的可以手動類比一下看與先序遍曆是否相同,然後從根結點一步步往下構造

class Solution {public:    void flatten(TreeNode *root) {        if(root == NULL) return;        while(root){            if(root->left){                TreeNode* ptr = root->left;                while(ptr->right) ptr = ptr->right;                ptr->right = root->right;                root->right = root->left;                root->left = NULL;            }            root = root->right;        }    }};