Leetcode 114. Flatten Binary Tree to Linked List

最新推荐文章于 2022-09-06 16:04:14 发布

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本文介绍了一种将二叉树结构通过算法转换为链表的方法，并提供了两种实现方式：一种是使用栈来辅助迭代操作，另一种是采用递归的方式进行节点连接。通过对这些方法的讲解，读者可以了解到如何在不借助额外数据结构的情况下，直接修改二叉树的左右子节点指针，从而将其转换成单链表。

Question

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

For example,
Given

The flattened tree should look like:

Code

public void flatten(TreeNode root) {
        if (root == null) {
            return;
        }
        TreeNode p = root;
        Stack<TreeNode> stacks = new Stack<>();
        if (root.right != null) {
            stacks.push(root.right);
        }
        if (root.left != null) {
            stacks.push(root.left);
        }
        while (!stacks.empty()) {
            TreeNode node = stacks.pop();
            p.right = node;
            p.left = null;
            p = p.right;
            if (node.right != null) {
                stacks.push(node.right);
            }
            if (node.left != null) {
                stacks.push(node.left);
            }
        }
    }

 /**
     * 递归的方式解决
     *
     * @param root
     * @return
     */
    // return : the tail of the list.
    public TreeNode dfs(TreeNode root) {
        if (root == null) {
            return null;
        }

        TreeNode left = root.left;
        TreeNode right = root.right;

        // Init the root.
        root.left = null;
        root.right = null;

        TreeNode tail = root;

        // connect the left tree.
        if (left != null) {
            tail.right = left;
            tail = dfs(left);
        }

        // connect the right tree.
        if (right != null) {
            tail.right = right;
            tail = dfs(right);
        }

        return tail;
    }