【重点】【二叉树】114. 二叉树展开为链表

题目

头插法

https://leetcode.cn/problems/flatten-binary-tree-to-linked-list/solutions/2992172/liang-chong-fang-fa-tou-cha-fa-fen-zhi-p-h9bg/?envType=study-plan-v2&envId=top-100-liked

Python

重点好法：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:

    def __init__(self):
        self.head = None

    def flatten(self, root: Optional[TreeNode]) -> None:
        """
        Do not return anything, modify root in-place instead.
        """
        if not root:
            return None
        self.flatten(root.right)
        self.flatten(root.left)
        root.left = None
        root.right = self.head
        self.head = root

法1：规律+迭代解法

参考答案很多：https://leetcode.cn/problems/flatten-binary-tree-to-linked-list/solutions/17274/xiang-xi-tong-su-de-si-lu-fen-xi-duo-jie-fa-by–26/?envType=study-plan-v2&envId=top-100-liked
重点记住下面这个！！！

class Solution {
    public void flatten(TreeNode root) {
        TreeNode cur = root;
        while (cur != null) {
            if (cur.left == null) { // 左子树为null直接下一个
                cur = cur.right;
            } else {
                TreeNode pre = cur.left; // 记录左子树的最右边节点
                while (pre.right != null) {
                    pre = pre.right;
                }
                pre.right = cur.right; // key1: 根节点的左子树放到<左子树-最右结点>的右子树
                cur.right = cur.left; // key2: 根结点的左子树变为右子树
                cur.left = null;      // 根结点左子树置null
                cur = cur.right;      // 下一个
            }
        }
    }
}

法2：先序遍历

思路不够好

// 递归版遍历
class Solution {
    public void flatten(TreeNode root) {
        List<TreeNode> nodeList = new ArrayList<>();
        preorder(root, nodeList);
        for (int i = 0; i < nodeList.size(); ++i) {
            TreeNode curNode = nodeList.get(i);
            curNode.left = null;
            if (i == nodeList.size() - 1) {
                curNode.right = null;
            } else {
                curNode.right = nodeList.get(i + 1);
            }
        }
    }
    
    public void preorder(TreeNode root, List<TreeNode> res) {
        if (root == null) {
            return;
        }
        res.add(root);
        preorder(root.left, res);
        preorder(root.right, res);
    }
}

// 迭代版遍历
class Solution {
    public void flatten(TreeNode root) {
        List<TreeNode> nodeList = new ArrayList<>();
        Stack<TreeNode> stack = new Stack<>();
        while (root != null || !stack.isEmpty()) {
            if (root != null) {
                nodeList.add(root);
                stack.push(root);
                root = root.left;
            } else {
                TreeNode tmp = stack.pop();
                root = tmp.right;
            }
        }
        for (int i = 0; i < nodeList.size(); ++i) {
            TreeNode curNode = nodeList.get(i);
            curNode.left = null;
            if (i == nodeList.size() - 1) {
                curNode.right = null;
            } else {
                curNode.right = nodeList.get(i + 1);
            }
        }
    }
}