51 lines
1.5 KiB
Java
51 lines
1.5 KiB
Java
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# 8. 二叉树的下一个结点
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[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)
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## 题目描述
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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。
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```java
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public class TreeLinkNode {
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int val;
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TreeLinkNode left = null;
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TreeLinkNode right = null;
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TreeLinkNode next = null;
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TreeLinkNode(int val) {
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this.val = val;
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}
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}
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```
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## 解题思路
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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；
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<img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/b0611f89-1e5f-4494-a795-3544bf65042a.gif" width="220px"/>
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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。
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<img src="https://cs-notes-1256109796.cos.ap-guangzhou.myqcloud.com/95080fae-de40-463d-a76e-783a0c677fec.gif" width="200px"/>
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```java
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public TreeLinkNode GetNext(TreeLinkNode pNode) {
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if (pNode.right != null) {
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TreeLinkNode node = pNode.right;
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while (node.left != null)
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node = node.left;
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return node;
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} else {
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while (pNode.next != null) {
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TreeLinkNode parent = pNode.next;
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if (parent.left == pNode)
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return parent;
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pNode = pNode.next;
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}
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}
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return null;
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}
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```
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