求一个二叉树中距离最远的两个节点

/*求二叉树中距离最远的两个点
 * 基本思路：
 * 递归计算两棵树的最大高度，设置一个全局变量，距离最远的两个节点element
 * 其中：在计算左子支，直接刷新上述全局变量，在计算右边子支时，设置两个临时Node变量，变量里的element用于
 * 保存右边子支的两个最远距离。根据比较两个距离的大小、其父节点所在的树三个的大小，来重新刷新全局变量。
 * 一个Trick~：在计算子支的最远距离的时候，因为要和其父节点所在的树比较大小，保存子支的最大距离的点数。 
 */


public class MaxLenTree {

    public Node root;

    public int len = 0;

    public class Node {
        char element;
        int hight;
        Node left;
        Node right;

        public Node(char element, int hight, Node left, Node right) {
            this.element = element;
            this.hight = hight;
            this.left = left;
            this.right = right;
        }

        Node() {

        }

    }

    MaxLenTree() {

/*        Node e = new Node('e', 0, null, null);
        Node d = new Node('d', 0, e, null);
        Node f = new Node('f', 0, null, null);
        Node c = new Node('c', 0, d, f);
        Node b = new Node('b', 0, c, null);
        Node a = new Node('A', 0, b, null);*/
        Node m = new Node('m', 0, null, null);
        Node n = new Node('n', 0, null, null);

        Node k = new Node('k', 0, m, n);
        Node l = new Node('l', 0, null, null);

        Node i = new Node('i', 0, null, null);
        Node j = new Node('j', 0, k, l);

        Node q = new Node('q', 0, null, null);
        Node p = new Node('P', 0, q, null);

        Node h = new Node('h', 0, p, null);
        Node d = new Node('d', 0, h, null);

        Node e = new Node('e', 0, i, j);
        Node b = new Node('B', 0, d, e);

        Node f = new Node('f', 0, null, null);
        Node g = new Node('g', 0, null, null);
        Node c = new Node('c', 0, f, g);

        Node a = new Node('A', 0, b, c);


        root = a;
    }

    public static void main(String[] args) {
        MaxLenTree mTree = new MaxLenTree();
        mTree.calculateHight();
        mTree.run();

    }

    public void run() {
        Node node = new Node();
        System.out.println(findMaxLen(this.root, node));
        System.out.println("maxNode left.element---" + maxNodeleft.element);
        System.out.println("maxNoderight.element---" + maxNoderight.element);

    }

    public Node maxNodeleft = new Node();
    public Node maxNoderight = new Node();

    public int findMaxLen(Node root, Node LongestChild) {
        System.out.println(LongestChild.element);
        if (root.left == null && root.right == null) {
            LongestChild.element = root.element;
            maxNodeleft.element = root.element;
            maxNoderight.element = root.element;
            return 0;

        }

        Node leftLongestChild = new Node();
        Node rightLogestChild = new Node();

        if (root.left == null || root.right == null) {
            if (root.right == null) {
                int a = findMaxLen(root.left, leftLongestChild);
                System.out.println(leftLongestChild.element);
                if (hight(root) > a) {
                    LongestChild.element = leftLongestChild.element;
                    this.maxNodeleft.element = leftLongestChild.element;
                    this.maxNoderight.element = root.element;
                    return hight(root);
                } else {

                    LongestChild.element = leftLongestChild.element;
                    return a;
                }
            } else {
                int a = findMaxLen(root.right, rightLogestChild);
                if (hight(root) > a) {
                    LongestChild.element = rightLogestChild.element;
                    this.maxNodeleft.element = root.element;
                    this.maxNoderight.element = rightLogestChild.element;
                    return hight(root);
                } else {
                    LongestChild.element = rightLogestChild.element;
                    return a;
                }
            }
        }

        int a = findMaxLen(root.left, leftLongestChild);

        Node rightNodetempLeft = new Node();
        rightNodetempLeft.element = maxNodeleft.element;
        Node rightNodetempRight = new Node();
        rightNodetempRight.element = maxNoderight.element;

        int b = findMaxLen(root.right, rightLogestChild);

        int c = Math.max(a, b);
        if (a > b) {
            maxNodeleft.element = rightNodetempLeft.element;
            maxNoderight.element = rightNodetempRight.element;
        }
        if (c == b) {
            LongestChild.element = rightLogestChild.element;
        } else {
            LongestChild.element = leftLongestChild.element;
        }
        int d = hight(root.left) + hight(root.right) + 2;

        if (c < d) {
            maxNodeleft.element = leftLongestChild.element;
            maxNoderight.element = rightLogestChild.element;
        }
        System.out.println(LongestChild.element);
        return Math.max(c, d);
    }


    public int hight(Node root) {
        if (root == null) {
            return -1;
        }
        return root.hight;
    }

    public void showTree(Node root) {
        if (root == null) {
            return;
        }
        System.out.println(root.element + "<----->" + root.hight);
        showTree(root.left);
        showTree(root.right);
    }

    public void calculateHight() {
        calculateHight(this.root);
    }

    public int calculateHight(Node root) {
        if (root == null) {
            return -1;
        }
        if (root.right == null && root.left == null) {
            root.hight = 0;
            return 0;
        }
        int temp = Math.max(calculateHight(root.left),
                calculateHight(root.right)) + 1;
        root.hight = temp;
        return temp;
    }

}

 

来源：https://www.cnblogs.com/lisongfeng9213/p/3806484.html