【Leetcode】116. 填充同一层的兄弟节点2

题目

给定一个完美二叉树，其所有叶子节点都在同一层，每个父节点都有两个子节点。二叉树定义如下：

struct Node {

int val;

Node *left;

Node *right;

Node *next;

}

填充它的每个 next 指针，让这个指针指向其下一个右侧节点。如果找不到下一个右侧节点，则将 next 指针设置为 NULL。

初始状态下，所有 next 指针都被设置为 NULL。

示例：

输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","left":null,"next":null,"right":null,"val":5},"val":2},"next":null,"right":{"$id":"5","left":{"$id":"6","left":null,"next":null,"right":null,"val":6},"next":null,"right":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}

输出：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":{"$id":"4","left":null,"next":{"$id":"5","left":null,"next":{"$id":"6","left":null,"next":null,"right":null,"val":7},"right":null,"val":6},"right":null,"val":5},"right":null,"val":4},"next":{"$id":"7","left":{"$ref":"5"},"next":null,"right":{"$ref":"6"},"val":3},"right":{"$ref":"4"},"val":2},"next":null,"right":{"$ref":"7"},"val":1}

解释：给定二叉树如图 A 所示，你的函数应该填充它的每个 next 指针，以指向其下一个右侧节点，如图 B 所示。

题解

方法一: 层序遍历

使用层序遍历，遍历的时候把同层的节点连接起来；

class Solution {
    public Node connect(Node root) {
        if (root == null) return null;
        Queue<Node> queue = new LinkedList<>();
        queue.add(root);
        while (!queue.isEmpty()) {
            int size = queue.size();
            Node current = null;
            while (size > 0) {
                Node node = queue.poll();
                if (node.right != null) queue.add(node.right);
                if (node.left != null) queue.add(node.left);
                node.next = current;
                current = node;
                size--;
            }
        }
        return root;
    }
}

方法二：递归

递归的时候我们通常就分解为递归子问题和递归结束条件。

递归子问题

• 左右子树分别连起来

递归结束条件

• node == null, 直接返回
• node.left != null, 把left.next连到node.right
• node.right != null && node.next != null, 把node的right 连到 node.next的left。例如遍历到2这个节点,把5连接到6.

class Solution {
    public Node connect(Node root) {
        // o(1) space.
        if (root == null) return null;
        if (root.left != null) root.left.next = root.right;
        if (root.right != null && root.next != null) root.right.next = root.next.left;
        connect(root.left);
        connect(root.right);    
        return root;
    }
}

方法三: 层序遍历 o(1)空间复杂度

层序遍历我们之前用队列来做,但是有时候我们会要求层序遍历用常数的空间复杂度来解。这种方法最关键的地方在于理解 如何从上一层切换到下一层的 。dummy的作用用于记录上一层的第一个节点是谁,每当遍历完一层之后，切到下一层.

class Solution {
    public Node connect(Node root) {
        Node dummy = new Node(0);
        Node pre = dummy;
        Node currentRoot = root;
        while (currentRoot != null) {
            if (currentRoot.left != null) {
                pre.next = currentRoot.left;
                pre = pre.next;
            }
            if (currentRoot.right != null) {
                pre.next = currentRoot.right;
                pre = pre.next;
            }
            currentRoot = currentRoot.next;
            if (currentRoot == null) {
                // 切换层.
                pre = dummy;
                currentRoot = dummy.next;
                dummy.next      = null;
            }
        }
        return root;
    }
}

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