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Commit 8ffdfd8

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solve 116.填充每个节点的下一个右侧节点指针
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/*
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* @lc app=leetcode.cn id=116 lang=java
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*
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* [116] 填充每个节点的下一个右侧节点指针
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*
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* https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node/description/
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*
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* algorithms
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* Medium (59.25%)
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* Likes: 191
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* Dislikes: 0
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* Total Accepted: 37.4K
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* Total Submissions: 61.8K
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* Testcase Example: '[1,2,3,4,5,6,7]'
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*
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* 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:
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*
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* struct Node {
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* ⁠ int val;
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* ⁠ Node *left;
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* ⁠ Node *right;
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* ⁠ Node *next;
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* }
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*
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* 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。
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*
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* 初始状态下,所有 next 指针都被设置为 NULL。
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*
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*
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*
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* 示例:
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*
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*
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*
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*
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* 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":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}
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*
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*
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* 输出:{"$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}
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*
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* 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。
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*
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*
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*
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*
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* 提示:
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*
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*
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* 你只能使用常量级额外空间。
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* 使用递归解题也符合要求,本题中递归程序占用的栈空间不算做额外的空间复杂度。
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*
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*
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*/
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// @lc code=start
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/*
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// Definition for a Node.
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class Node {
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public int val;
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public Node left;
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public Node right;
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public Node next;
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public Node() {}
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public Node(int _val) {
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val = _val;
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}
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public Node(int _val, Node _left, Node _right, Node _next) {
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val = _val;
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left = _left;
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right = _right;
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next = _next;
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}
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};
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*/
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class Solution {
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public Node connect(Node root) {
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if (root == null) {
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return null;
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}
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Queue<Node> queue = new LinkedList<>();
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queue.offer(root);
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while (!queue.isEmpty()) {
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int size = queue.size();
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Node pre = null;
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for (int i = 0; i < size; i++) {
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Node node = queue.poll();
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node.next = pre;
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pre = node;
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if (node.right != null) {
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queue.offer(node.right);
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}
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if (node.left != null) {
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queue.offer(node.left);
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}
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}
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}
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return root;
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}
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}
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// @lc code=end
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/*
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* @lc app=leetcode.cn id=117 lang=java
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*
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* [117] 填充每个节点的下一个右侧节点指针 II
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*
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* https://leetcode-cn.com/problems/populating-next-right-pointers-in-each-node-ii/description/
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*
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* algorithms
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* Medium (48.62%)
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* Likes: 162
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* Dislikes: 0
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* Total Accepted: 23.2K
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* Total Submissions: 46.7K
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* Testcase Example: '[1,2,3,4,5,null,7]'
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*
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* 给定一个二叉树
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*
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* struct Node {
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* ⁠ int val;
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* ⁠ Node *left;
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* ⁠ Node *right;
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* ⁠ Node *next;
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* }
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*
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* 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。
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*
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* 初始状态下,所有 next 指针都被设置为 NULL。
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*
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*
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*
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* 进阶:
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*
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*
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* 你只能使用常量级额外空间。
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* 使用递归解题也符合要求,本题中递归程序占用的栈空间不算做额外的空间复杂度。
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*
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*
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*
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*
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* 示例:
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*
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*
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*
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* 输入:root = [1,2,3,4,5,null,7]
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* 输出:[1,#,2,3,#,4,5,7,#]
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* 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。
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*
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*
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*
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* 提示:
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*
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*
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* 树中的节点数小于 6000
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* -100 <= node.val <= 100
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*
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*
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*
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*
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*
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*
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*
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*/
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// @lc code=start
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/*
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// Definition for a Node.
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class Node {
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public int val;
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public Node left;
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public Node right;
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public Node next;
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public Node() {}
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public Node(int _val) {
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val = _val;
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}
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public Node(int _val, Node _left, Node _right, Node _next) {
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val = _val;
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left = _left;
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right = _right;
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next = _next;
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}
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};
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*/
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class Solution {
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public Node connect(Node root) {
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if (root == null) {
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return null;
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}
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Queue<Node> queue = new LinkedList<>();
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queue.offer(root);
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while (!queue.isEmpty()) {
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int size = queue.size();
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Node pre = null;
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for (int i = 0; i < size; i++) {
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Node node = queue.poll();
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node.next = pre;
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pre = node;
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if (node.right != null) {
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queue.offer(node.right);
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}
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if (node.left != null) {
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queue.offer(node.left);
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}
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}
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}
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return root;
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}
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}
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// @lc code=end
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