114. Flatten Binary Tree to Linked List二叉树展开为链表

114 Flatten Binary Tree to Linked List 原题链接

递归dfs搜索思想

= = 题意是要就地开花，改那么一下，我个憨憨自己搞出了个新树，看了题解才知道，害。

旧版代码 （没过）

/**
 * 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:
    TreeNode* tmp = nullptr;
public:
    void flatten(TreeNode* root) {
        if(!root) return;
        tmp = new TreeNode(root->val);       
        TreeNode* head = nullptr;
        head = tmp;
        dfs(root);
        delete(root);
        root = head;
        /*
        cout<<root->val;
        while(root->right){
            cout<<root->right->val<<endl;
            root = root->right;
        }
        */
    }
    void dfs(TreeNode* root){
        if(root->left){
            tmp->right = new TreeNode(root->left->val);
            tmp = tmp->right;
           // cout<<tmp->val<<endl;
            dfs(root->left);
        }
        if(root->right){
            tmp->right = new TreeNode(root->right->val);
            tmp = tmp->right;
            //cout<<tmp->val<<endl;
            dfs(root->right);
        }
    }
};

船新版本

class Solution {
public:
    void flatten(TreeNode* root) {
        if(!root) return;
        vector<TreeNode*> node;
        dfs(root,node);
        for(int i=0;i<node.size()-1;i++){
            node[i]->left=NULL;
            node[i]->right=node[i+1];
        }
    }
    void dfs(TreeNode* root, vector<TreeNode*> &node){
        if(!root){
            return;
        }
        node.push_back(root);
        dfs(root->left,node);
        dfs(root->right,node);
    }
};