Clear and Concise C++ Solution


  • 0
    C

    We do this recursively.
    for a node root, we flatten its left node and right node:

    • Using tail to record the left node's linklist's tail
    • Using tailtail to record the right node's linklist's tail
      then we connect tail and root->right, root and root->left.
      The code is below:
    /**
     * 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* dfs(TreeNode* root) {
            if (!root) return NULL;
            TreeNode* tail = dfs(root->left);
            TreeNode* tailtail = dfs(root->right);
            if (tail) {
                tail->right = root->right;
                root->right = root->left;
                root->left = NULL;
            }
            return tailtail ? tailtail : (tail ? tail : root);
        }
        
        void flatten(TreeNode* root) {
            if (!root) return;
            dfs(root);
        }
    };
    

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