58 lines
1.7 KiB
Java
58 lines
1.7 KiB
Java
|
# 8. 二叉树的下一个结点
|
|||
|
|
|
|||
|
|
[NowCoder](https://www.nowcoder.com/practice/9023a0c988684a53960365b889ceaf5e?tpId=13&tqId=11210&tPage=1&rp=1&ru=/ta/coding-interviews&qru=/ta/coding-interviews/question-ranking&from=cyc_github)
|
|||
|
|
|
|||
|
|
## 题目描述
|
|||
|
|
|
|||
|
|
给定一个二叉树和其中的一个结点，请找出中序遍历顺序的下一个结点并且返回。注意，树中的结点不仅包含左右子结点，同时包含指向父结点的指针。
|
|||
|
|
|
|||
|
|
```java
|
|||
|
|
public class TreeLinkNode {
|
|||
|
|
|
|||
|
|
int val;
|
|||
|
|
TreeLinkNode left = null;
|
|||
|
|
TreeLinkNode right = null;
|
|||
|
|
TreeLinkNode next = null;
|
|||
|
|
|
|||
|
|
TreeLinkNode(int val) {
|
|||
|
|
this.val = val;
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
## 解题思路
|
|||
|
|
|
|||
|
|
① 如果一个节点的右子树不为空，那么该节点的下一个节点是右子树的最左节点；
|
|||
|
|
|
|||
|
|
<div align="center"> <img src="pics/b0611f89-1e5f-4494-a795-3544bf65042a.gif" width="220px"/> </div><br>
|
|||
|
|
|
|||
|
|
② 否则，向上找第一个左链接指向的树包含该节点的祖先节点。
|
|||
|
|
|
|||
|
|
<div align="center"> <img src="pics/95080fae-de40-463d-a76e-783a0c677fec.gif" width="200px"/> </div><br>
|
|||
|
|
|
|||
|
|
```java
|
|||
|
|
public TreeLinkNode GetNext(TreeLinkNode pNode) {
|
|||
|
|
if (pNode.right != null) {
|
|||
|
|
TreeLinkNode node = pNode.right;
|
|||
|
|
while (node.left != null)
|
|||
|
|
node = node.left;
|
|||
|
|
return node;
|
|||
|
|
} else {
|
|||
|
|
while (pNode.next != null) {
|
|||
|
|
TreeLinkNode parent = pNode.next;
|
|||
|
|
if (parent.left == pNode)
|
|||
|
|
return parent;
|
|||
|
|
pNode = pNode.next;
|
|||
|
|
}
|
|||
|
|
}
|
|||
|
|
return null;
|
|||
|
|
}
|
|||
|
|
```
|
|||
|
|
|
|||
|
|
|
|||
|
|
|
|||
|
|
|
|||
|
|
|
|||
|
|
|
|||
|
|
<div align="center"><img width="320px" src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/githubio/公众号二维码-1.png"></img></div>
|