114. Flatten Binary Tree to Linked List¶

114. Flatten Binary Tree to Linked List¶

Tags: Medium Tree Depth-first Search

Links: https://leetcode.com/problems/flatten-binary-tree-to-linked-list/

Given a binary tree, flatten it to a linked list in-place.

For example, given the following tree:

The flattened tree should look like:

/**
 * 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) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        if (!root) return;

        if (root -> left) flatten(root -> left);
        if (root -> right) flatten(root -> right);

        TreeNode *tmp = root -> right;
        root -> right = root -> left;
        root -> left = nullptr;

        TreeNode *p = root;
        while (p -> right) p = p -> right;
        p -> right = tmp;
    }
};

递归写法，先让根节点的左右子树达成目标，然后就是链表的操作。

/**
 * 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) {
        std::ios_base::sync_with_stdio(false);
        cin.tie(NULL);
        cout.tie(NULL);

        TreeNode * cur = root;
        while (cur) {
            if (cur -> left) {
                TreeNode * p = cur -> left;
                while (p -> right) p = p -> right;
                p -> right = cur -> right;
                cur -> right = cur -> left;
                cur -> left = nullptr;
            }
            cur = cur -> right;
        }
    }
};

从根节点开始出发，先检测其左子结点是否存在，如存在则将根节点和其右子节点断开，将左子结点及其后面所有结构一起连到原右子节点的位置，把原右子节点连到元左子结点最后面的右子节点之后。

例如，对于下面的二叉树，上述算法的变换的过程如下：