分析：

暴力求每一段距离也可。

 

对于以本节点为根的二叉树，最远距离有三种可能：

1）最远路径来自左子树

2 ）最远路径来自右子树（图示与左子树同理）

3）最远路径为左右子树距离根最远的两个节点，经过根结点连起来。

（多种最长路径）

 

需要的信息：

1）左子树的最远路径长度

2）右子树的最远路径长度

3）左右子树的深度（深度即最远节点）

 

定义结点：

    public static class Node {public int value;public Node left;public Node right;public Node(int data) {this.value = data;}}

构造返回值信息：

    public static class ReturnType{public int maxDistance;//最长距离public int h;          //高度public ReturnType(int m, int h) {this.maxDistance = m;;this.h = h;}}

求解过程比较好写了：

	public static ReturnType process(Node head) {if(head == null) {return new ReturnType(0,0);}//收信息ReturnType leftReturnType = process(head.left);ReturnType rightReturnType = process(head.right);int includeHeadDistance = leftReturnType.h + 1 + rightReturnType.h;//情况3int p1 = leftReturnType.maxDistance;int p2 = rightReturnType.maxDistance;int resultDistance = Math.max(Math.max(p1, p2), includeHeadDistance);//最长距离int hitself  = Math.max(leftReturnType.h, leftReturnType.h) + 1;     //树的高度等于子树高度+1return new ReturnType(resultDistance, hitself);}

优化：

其实我们不需要返回左右子树深度，可以用一个全局变量记录。遇到NULL把变量记为0，不影响接下来的计算。

用一个含一个元素的数组来记录。因为某些二叉树题目不只需要一个信息，所以要利用全局数组。

	public static int posOrder(Node head, int[] record) {if (head == null) {record[0] = 0;//重要return 0;}//取信息int lMax = posOrder(head.left, record);int maxfromLeft = record[0];int rMax = posOrder(head.right, record);int maxFromRight = record[0];int curNodeMax = maxfromLeft + maxFromRight + 1;//情况3record[0] = Math.max(maxfromLeft, maxFromRight) + 1;return Math.max(Math.max(lMax, rMax), curNodeMax);}

最后放上全部代码：

package q;public class Demo {public static class Node {public int value;public Node left;public Node right;public Node(int data) {this.value = data;}}public static int maxDistance(Node head) {int[] record = new int[1];return posOrder(head, record);}public static class ReturnType{public int maxDistance;//最长距离public int h;          //高度public ReturnType(int m, int h) {this.maxDistance = m;;this.h = h;}}public static ReturnType process(Node head) {if(head == null) {return new ReturnType(0,0);}//收信息ReturnType leftReturnType = process(head.left);ReturnType rightReturnType = process(head.right);int includeHeadDistance = leftReturnType.h + 1 + rightReturnType.h;//情况3int p1 = leftReturnType.maxDistance;int p2 = rightReturnType.maxDistance;int resultDistance = Math.max(Math.max(p1, p2), includeHeadDistance);//最长距离int hitself  = Math.max(leftReturnType.h, leftReturnType.h) + 1;     //树的高度等于子树高度+1return new ReturnType(resultDistance, hitself);}public static int posOrder(Node head, int[] record) {if (head == null) {record[0] = 0;//重要return 0;}//取信息int lMax = posOrder(head.left, record);int maxfromLeft = record[0];int rMax = posOrder(head.right, record);int maxFromRight = record[0];int curNodeMax = maxfromLeft + maxFromRight + 1;//情况3record[0] = Math.max(maxfromLeft, maxFromRight) + 1;return Math.max(Math.max(lMax, rMax), curNodeMax);}public static void main(String[] args) {Node head1 = new Node(1);head1.left = new Node(2);head1.right = new Node(3);head1.left.left = new Node(4);head1.left.right = new Node(5);head1.right.left = new Node(6);head1.right.right = new Node(7);head1.left.left.left = new Node(8);head1.right.left.right = new Node(9);System.out.println(maxDistance(head1));Node head2 = new Node(1);head2.left = new Node(2);head2.right = new Node(3);head2.right.left = new Node(4);head2.right.right = new Node(5);head2.right.left.left = new Node(6);head2.right.right.right = new Node(7);head2.right.left.left.left = new Node(8);head2.right.right.right.right = new Node(9);System.out.println(maxDistance(head2));}}