二叉树最长路径

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3
		int 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+1
		
		return 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3
		record[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3
		int 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+1
		
		return 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3
		record[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));

	}

}