# 二叉树的下一个结点

## 题目：给定一棵二叉树和其中的一个结点，如何找出中序遍历顺序的下一个结点？树中的结点除了有两个分别指向左右子结点的指针以外，还有一个指向父节点的指针。

### 结点定义

private static class BinaryTreeNode {
private int val;
private BinaryTreeNode left;
private BinaryTreeNode right;
private BinaryTreeNode parent;
public BinaryTreeNode() {
}
public BinaryTreeNode(int val) {
this.val = val;
}
@Override
public String toString() {
return val + "";
}
}

### 代码实现

public class Test58 {
private static class BinaryTreeNode {
private int val;
private BinaryTreeNode left;
private BinaryTreeNode right;
private BinaryTreeNode parent;
public BinaryTreeNode() {
}
public BinaryTreeNode(int val) {
this.val = val;
}
@Override
public String toString() {
return val + "";
}
}
public static BinaryTreeNode getNext(BinaryTreeNode node) {
if (node == null) {
return null;
}
// 保存要查找的下一个节点
BinaryTreeNode target = null;
if (node.right != null) {
target = node.right;
while (target.left != null) {
target = target.left;
}
return target;
} else if (node.parent != null){
target = node.parent;
BinaryTreeNode cur = node;
// 如果父新结点不为空，并且，子结点不是父结点的左孩子
while (target != null && target.left != cur) {
cur = target;
target = target.parent;
}
return target;
}
return null;
}
private static void assemble(BinaryTreeNode node,
BinaryTreeNode left,
BinaryTreeNode right,
BinaryTreeNode parent) {
node.left = left;
node.right = right;
node.parent = parent;
}
public static void main(String[] args) {
test01();
}
//                            1
//                  2                   3
//             4         5          6          7
//          8     9   10   11   12   13    14   15
public static void test01() {
BinaryTreeNode n1 = new BinaryTreeNode(1); // 12
BinaryTreeNode n2 = new BinaryTreeNode(2); // 10
BinaryTreeNode n3 = new BinaryTreeNode(3); // 14
BinaryTreeNode n4 = new BinaryTreeNode(4); // 9
BinaryTreeNode n5 = new BinaryTreeNode(5); // 11
BinaryTreeNode n6 = new BinaryTreeNode(6); // 13
BinaryTreeNode n7 = new BinaryTreeNode(7); // 15
BinaryTreeNode n8 = new BinaryTreeNode(8); // 4
BinaryTreeNode n9 = new BinaryTreeNode(9); // 2
BinaryTreeNode n10 = new BinaryTreeNode(10); // 5
BinaryTreeNode n11 = new BinaryTreeNode(11); // 1
BinaryTreeNode n12 = new BinaryTreeNode(12); // 6
BinaryTreeNode n13 = new BinaryTreeNode(13); // 3
BinaryTreeNode n14 = new BinaryTreeNode(14); // 7
BinaryTreeNode n15 = new BinaryTreeNode(15); // null
assemble(n1, n2, n3, null);
assemble(n2, n4, n5, n1);
assemble(n3, n6, n7, n1);
assemble(n4, n8, n9, n2);
assemble(n5, n10, n11, n2);
assemble(n6, n12, n13, n3);
assemble(n7, n14, n15, n3);
assemble(n8, null, null, n4);
assemble(n9, null, null, n4);
assemble(n10, null, null, n5);
assemble(n11, null, null, n5);
assemble(n12, null, null, n6);
assemble(n13, null, null, n6);
assemble(n14, null, null, n7);
assemble(n15, null, null, n7);
System.out.println(getNext(n1));
System.out.println(getNext(n2));
System.out.println(getNext(n3));
System.out.println(getNext(n4));
System.out.println(getNext(n5));
System.out.println(getNext(n6));
System.out.println(getNext(n7));
System.out.println(getNext(n8));
System.out.println(getNext(n9));
System.out.println(getNext(n10));
System.out.println(getNext(n11));
System.out.println(getNext(n12));
System.out.println(getNext(n13));
System.out.println(getNext(n14));
System.out.println(getNext(n15));
}
}