找二叉树中一个节点的后继节点，前驱节点

后继节点：后继节点是二叉树中序遍历后的集合每个节点的下一个节点

前驱节点：二叉树中序遍历后的集合每个节点的前一个节点

问：给定某个节点，怎么找到他的后继节点？

        如何避免遍历整棵树生成整个序列，能不能通过这个节点与它后继节点的距离，来找到某个节点的后继节点？

规律是这样的：假定某个节点X，

情况一：如果X有右子树，则节点X的后继节点一定是它的右子树的最左的节点。（因为中序遍历的顺序是左，中，右，则节点X                   打印之后，就该打印它的右子树了，而打印它的右子树肯定最先打印右子树的最左节点了）

情况二：如果X没有右子树，就要考虑X是作为哪个节点的左子树的最后一个节点？也就是说哪个节点的左子树是以X结尾的？则                这个节点就是X的后继节点。也就是说X没有右子树就通过它的parent指针往上找，直到找到当前节点是它父节点的左孩                子的时候停下来，则这个父节点就是要找的X的后继节点。

package com.tree;

public class SuccessorNode {

	public static class Node {
		public int value;
		public Node left;
		public Node right;
		public Node parent;

		public Node(int data) {
			this.value = data;
		}
	}

	public static Node getSuccessorNode(Node node) {
		if (node == null) {  
			return node;
		}
		if (node.right != null) { //如果当前节点的右孩子节点不为空，说明有右子树，
			return getLeftMost(node.right); //则找到并返回右子树上最左的节点
		} else {               //如果当前节点没有右子树
			Node parent = node.parent;
			while (parent != null && parent.left != node) {
				node = parent;
				parent = node.parent;
			}
			return parent;
		}
	}

	public static Node getLeftMost(Node node) { //在这个函数里面，node是某个节点的头部
		if (node == null) {
			return node;
		}
		while (node.left != null) { //左子树不为空的情况下，一路向左
			node = node.left;
		}
		return node;
	}

	public static void main(String[] args) {
		Node head = new Node(6);
		head.parent = null;
		head.left = new Node(3);
		head.left.parent = head;
		head.left.left = new Node(1);
		head.left.left.parent = head.left;
		head.left.left.right = new Node(2);
		head.left.left.right.parent = head.left.left;
		head.left.right = new Node(4);
		head.left.right.parent = head.left;
		head.left.right.right = new Node(5);
		head.left.right.right.parent = head.left.right;
		head.right = new Node(9);
		head.right.parent = head;
		head.right.left = new Node(8);
		head.right.left.parent = head.right;
		head.right.left.left = new Node(7);
		head.right.left.left.parent = head.right.left;
		head.right.right = new Node(10);
		head.right.right.parent = head.right;

		Node test = head.left.left;
		System.out.println(test.value + " next: " + getSuccessorNode(test).value);
		test = head.left.left.right;
		System.out.println(test.value + " next: " + getSuccessorNode(test).value);
		test = head.left;
		System.out.println(test.value + " next: " + getSuccessorNode(test).value);
		test = head.left.right;
		System.out.println(test.value + " next: " + getSuccessorNode(test).value);
		test = head.left.right.right;
		System.out.println(test.value + " next: " + getSuccessorNode(test).value);
		test = head;
		System.out.println(test.value + " next: " + getSuccessorNode(test).value);
		test = head.right.left.left;
		System.out.println(test.value + " next: " + getSuccessorNode(test).value);
		test = head.right.left;
		System.out.println(test.value + " next: " + getSuccessorNode(test).value);
		test = head.right;
		System.out.println(test.value + " next: " + getSuccessorNode(test).value);
		test = head.right.right; // 10's next is null
		System.out.println(test.value + " next: " + getSuccessorNode(test));
	}

}

代码测试结果：

那么同样的，X的前驱节点怎么找？（和后继节点是对应的规律）

1.如果X有左子树，那么前驱节点是X的左子树上的最右节点

2.如果X没有左子树，那么同样往上找，直到当前节点是他父节点的右孩子的时候停下来，这个父节点就是X的前驱节点