专栏首页米扑专栏【leetcode】Flatten Binary Tree to Linked List

【leetcode】Flatten Binary Tree to Linked List

Question : 

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

click to show hints.

Anwser 1 :

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root == NULL) return;

        TreeNode* left = root->left;
        TreeNode* right = root->right;      // the original right child

        if (left) {
            root->right = left;
            root->left = NULL;

            TreeNode* rightmost = left;
            while(rightmost->right) {
                rightmost = rightmost->right;
                
            }
            rightmost->right = right; // point to the original right child
        }

        flatten(root->right);    
    }
};

Anwser 2 :

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        if (root == NULL) return;
        
        if (root->left)
            flatten(root->left);
            
        if (root->right)
            flatten(root->right);
            
        if (NULL == root->left)
            return;
            
        TreeNode ** ptn = &(root->left->right);
        while (*ptn) {
            ptn = & ((*ptn)->right);
        }
            
        *ptn = root->right;
        root->right = root->left;   // link right to left
        root->left = NULL;
    }
};

Anwser 3 :  

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        bool flag = true;
        stack<TreeNode *> t;
        TreeNode *pre = NULL;
        
        if(root) {
            t.push(root);
        }
        
        while(t.size()) {
            TreeNode *cur = t.top();
            if(flag) {
                if(cur->left && cur->left != pre) {
                    t.push(cur->left);
                } else {
                    flag = false;
                }
            } else {
                if(cur->right && cur->right != pre) {
                    t.push(cur->right);
                    flag = true;
                } else {
                    t.pop();
                    //at this time, the sub-tree of cur is flattened, so just flatten the cur
                    TreeNode *left = cur->left;
                    if(left) {
                        TreeNode *lastLeft = left;
                        while(lastLeft->right) {
                            lastLeft = lastLeft->right;
                        }
                        lastLeft->right = cur->right;
                        cur->right = left;
                        cur->left = NULL;
                    }
                }
            }
            pre = cur;
        }
    }
};

参考推荐:

LeetCode Problem:Flatten Binary Tree to Linked List

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