116. 填充每個節(jié)點的下一個右側(cè)節(jié)點指針

難度中等128

給定一個完美二叉樹，其所有葉子節(jié)點都在同一層，每個父節(jié)點都有兩個子節(jié)點。二叉樹定義如下：

struct Node {int val;Node *left;Node *right;Node *next; }

填充它的每個 next 指針，讓這個指針指向其下一個右側(cè)節(jié)點。如果找不到下一個右側(cè)節(jié)點，則將 next 指針設(shè)置為?NULL。

初始狀態(tài)下，所有?next 指針都被設(shè)置為?NULL。

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示例：

輸入：{"$id":"1","left":{"$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":{"$id":"7","left":null,"next":null,"right":null,"val":7},"val":3},"val":1}輸出：{"$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}解釋：給定二叉樹如圖 A 所示，你的函數(shù)應(yīng)該填充它的每個 next 指針，以指向其下一個右側(cè)節(jié)點，如圖 B 所示。

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提示：

• 你只能使用常量級額外空間。
• 使用遞歸解題也符合要求，本題中遞歸程序占用的棧空間不算做額外的空間復(fù)雜度。

思路：層序遍歷稍微改一下，把相同層的連一下即可。

/* // Definition for a Node. class Node {public int val;public Node left;public Node right;public Node next;public Node() {}public Node(int _val) {val = _val;}public Node(int _val, Node _left, Node _right, Node _next) {val = _val;left = _left;right = _right;next = _next;} }; */ class Solution {public Node connect(Node root) {if (root == null) {return null;}Queue<Node> Q = new LinkedList<Node>(); Q.add(root);while (Q.size() > 0) {int size = Q.size();for(int i = 0; i < size; i++) {Node node = Q.poll();if (i < size - 1) {node.next = Q.peek();}if (node.left != null) {Q.add(node.left);}if (node.right != null) {Q.add(node.right);}}}return root;} }

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總結(jié)

以上是生活随笔為你收集整理的leetcode116. 填充每个节点的下一个右侧节点指针的全部內(nèi)容，希望文章能夠幫你解決所遇到的問題。

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