[LeetCode] 114. Flatten Binary Tree to Linked List

Given the root of a binary tree, flatten the tree into a "linked list":

• The "linked list" should use the same TreeNode class where the right child pointer points to the next node in the list and the left child pointer is always null.
• The "linked list" should be in the same order as a pre-order traversal of the binary tree.

Example 1:

Input: root = [1,2,5,3,4,null,6]
Output: [1,null,2,null,3,null,4,null,5,null,6]

Example 2:
Input: root = []
Output: []

Example 3:
Input: root = [0]
Output: [0]

Constraints:
The number of nodes in the tree is in the range [0, 2000].
-100 <= Node.val <= 100

Follow up: Can you flatten the tree in-place (with O(1) extra space)?

二叉树展开为链表。

给你二叉树的根结点 root ，请你将它展开为一个单链表：

展开后的单链表应该同样使用 TreeNode ，其中 right 子指针指向链表中下一个结点，而左子指针始终为 null 。
展开后的单链表应该与二叉树 先序遍历 顺序相同。

来源：力扣（LeetCode）
链接：https://leetcode-cn.com/problems/flatten-binary-tree-to-linked-list
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思路一 - 迭代

这里我给出迭代的做法，会用到 stack。照着例子看来，最后的输出是按照先序遍历的顺序来的。所以用 stack 先右后左地塞入每个节点，但是在弹出的时候需要注意一些细节，在重连 right 指针的时候，先不要把那个对应的右节点从 stack 中 pop 出来，否则就会出错。具体的参见代码注释。

复杂度

时间O(n)
空间O(n) - stack

代码

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public void flatten(TreeNode root) {
        // corner case
        if (root == null) {
            return;
        }

        // normal case
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        while (!stack.isEmpty()) {
            TreeNode cur = stack.pop();
            if (cur.right != null) {
                stack.push(cur.right);
            }
            if (cur.left != null) {
                stack.push(cur.left);
            }
            if (!stack.isEmpty()) {
                // if pop here, the result will be wrong
                // at this step, you get the right node, this node will be poped out in the next round
                cur.right = stack.peek();
            }
            cur.left = null;
        }
    }
}

JavaScript实现

/**
 * @param {TreeNode} root
 * @return {void} Do not return anything, modify root in-place instead.
 */
var flatten = function (root) {
    // corner case
    if (root == null) {
        return;
    }

    // normal case
    let stack = [];
    stack.push(root);
    while (stack.length) {
        let cur = stack.pop();
        if (cur.right != null) {
            stack.push(cur.right);
        }
        if (cur.left != null) {
            stack.push(cur.left);
        }
        if (stack.length) {
            cur.right = stack[stack.length - 1];
        }
        cur.left = null;
    }
};

思路二 - 也是迭代，无需额外空间

参考这个帖子的解法一。

复杂度

时间O(n)
空间O(n) - stack

代码

Java实现

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public void flatten(TreeNode root) {
        while (root != null) {
            // 如果左子树为空，直接去看右子树
            if (root.left == null) {
                root = root.right;
            } else {
                // 找到左子树的最右孩子，走到4的位置
                TreeNode pre = root.left;
                while (pre.right != null) {
                    pre = pre.right;
                }
                // 把左子树的最右孩子连接到右子树
                pre.right = root.right;        // 4 - 5
                root.right = root.left;        // 1 - 2，只是2从1的左孩子变成了右孩子
                root.left = null;            // 1的左指针设置为NULL
                root = root.right;            // 1走到2
            }
        }
    }
}

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