LeetCode 114: Flatten Binary Tree to Linked List

Flatten Binary Tree to Linked List

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

Hints:
If you notice carefully in the flattened tree, each node’s right child points to the next node of a pre-order traversal.

解题思路

思路一：变换后的树实际上是按照先序遍历的顺序排列的，因此我们可以在先序遍历的过程中记录先序遍历的前驱结点，调整前驱结点的左右子树。这里需要注意的是遍历过程中需要先将当前节点的右子树记录下来，再递归调整当前节点的左右子树。代码如下：

/**
 * 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 {
private:
    void preorderFlatten(TreeNode *root, TreeNode *&pre) {
        if (root == NULL) return;

        // pre 保存前序遍历序列中的前一个节点
        if (pre) {
            pre->left = NULL;
            pre->right = root;
        }
        pre = root;

        TreeNode *right = root->right; // Note：先把右子树保存下来

        preorderFlatten(root->left, pre); //递归
        preorderFlatten(right, pre); // 递归
    }
public:
    void flatten(TreeNode* root) {
        TreeNode *pre = NULL;
        return preorderFlatten(root, pre);
    }
};

思路二：从根节点开始，如果当前节点有左子树，就将它的左子树添加到自己和右子树之间：既将当前节点的右子树作为左子树最右边的节点的右子树。然后依次往右处理当前结点的右子树，直到右子树为空。代码如下：

/**
 * 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:
    void flatten(TreeNode* root) {
        if (root == NULL) return;

        while(root != NULL) {
            if (root->left != NULL) {
                TreeNode *pre = root->left; // pre 指向左子树最右边的节点

                while (pre->right != NULL) pre = pre->right;

                pre->right = root->right;
                root->right = root->left;
                root->left = NULL;
            }
            root = root->right;
        }
    }
};