每天做一道算法題,循序漸進,按算法分類刷題。堅持下去,看能堅持多久,也看最終能有多大成效。
填充每個節點的下一個右側節點指針
給定一個完美二叉樹,其所有葉子節點都在同一層,每個父節點都有兩個子節點。二叉樹定義如下:
<code>struct Node {
int val;
Node *left;
Node *right;
Node *next;
}/<code>
填充它的每個 next 指針,讓這個指針指向其下一個右側節點。如果找不到下一個右側節點,則將 next 指針設置為 NULL。
初始狀態下,所有 next 指針都被設置為 NULL。
示例:
![每日算法練習20200404](http://p2.ttnews.xyz/loading.gif)
<code>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,你的函數應該填充它的每個 next 指針,以指向其下一個右側節點,如圖 B 所示。
/<code>
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,本題中遞歸程序佔用的棧空間不算做額外的空間複雜度。
解決方案
使用遞歸來實現,當左子節點不為null時,把左子節點的next指針指向右子節點。噹噹前節點的next指針不為null時,將右子節點的next指針指向當前節點next指針的左子節點。
實現代碼
![每日算法練習20200404](http://p2.ttnews.xyz/loading.gif)
鏈接指針
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