【96.二叉树展开为链表】

题目

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。

示例 1：

输入： root = [1,2,5,3,4,null,6]
输出： [1,null,2,null,3,null,4,null,5,null,6]

题解

方式一：前序遍历（递归）

Deque<TreeNode> deque = new LinkedList<>();

public void flatten(TreeNode root) {
    TreeNode cur = root;
    preOrder(root);
    deque.pollFirst();
    // 重新构建二叉树
    while (!deque.isEmpty()) {
        cur.right = deque.pollFirst();
        cur = cur.right;
    }
}

public void preOrder(TreeNode node) {
    if (node == null) {
        return;
    }
    deque.offerLast(node);
    // 断开原来的节点
    if (node.left != null) {
        preOrder(node.left);
        node.left = null;
    }
    if (node.right != null) {
        preOrder(node.right);
        node.right = null;
    }
}

方式二：前序遍历（迭代）

public void flatten(TreeNode root) {
    List<TreeNode> list = new ArrayList<>(); // 本题需要记录全序遍历节点
    Stack<TreeNode> stack = new Stack<>();
    TreeNode node = root;
    while (node != null || !stack.isEmpty()) {
        while (node != null) {
            list.add(node);
            stack.push(node);
            node = node.left;
        }
        node = stack.pop();
        node = node.right;
    }
    for (int i = 1; i < list.size(); i++) { // 展开
        TreeNode pre = list.get(i - 1);
        TreeNode cur = list.get(i);
        pre.left = null;
        pre.right = cur;
    }
}

方式三：前序遍历

public void flatten(TreeNode root) {
    if(root == null) {
        return;
    }

    Stack<TreeNode> stack = new Stack<>();
    stack.push(root);
    TreeNode pre = null;
    while (!stack.isEmpty()) {
        TreeNode cur = stack.pop();
        if (pre != null) {
            pre.left = null;
            pre.right = cur;
        }
        TreeNode left = cur.left, right = cur.right;
        if (right != null) {
            stack.push(right);
        }
        if (left != null) {
            stack.push(left);
        }
        pre = cur;
    }
}

方式四：递归

public void flatten(TreeNode root) {
    if(root == null) {
        return;
    }

    flatten(root.left);
    flatten(root.right);

    TreeNode tem = root.right;
    root.right = root.left;
    root.left = null;
    while (root.right != null) {
        root = root.right;
    }
    root.right = tem;
}

总结

算法：递归、前序遍历