Leetcode Flatten Binary Tree to Linked List

OpenSoucre發表於2014-07-02

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

此題看題目給出的示例就知道是先序遍歷
本題用遞迴求先序遍歷
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,由於題目是先序遍歷,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模擬一下看與先序遍歷是否相同,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;
        }
    }
};

 


 

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