剑指Offer-58

来源：互联网发布：盈建科软件破解版编辑：程序博客网时间：2024/05/17 23:33
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。
实现

public class Solution58 {    public static BinaryTreeNode2 getNext(BinaryTreeNode2 node) {        if (node == null) {            return null;        }        // 保存要查找的下一个节点        BinaryTreeNode2 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;            BinaryTreeNode2 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;    }    //构建树    private static void assemble(BinaryTreeNode2 node,                                 BinaryTreeNode2 left,                                 BinaryTreeNode2 right,                                 BinaryTreeNode2 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() {        BinaryTreeNode2 n1 = new BinaryTreeNode2(1); // 12        BinaryTreeNode2 n2 = new BinaryTreeNode2(2); // 10        BinaryTreeNode2 n3 = new BinaryTreeNode2(3); // 14        BinaryTreeNode2 n4 = new BinaryTreeNode2(4); // 9        BinaryTreeNode2 n5 = new BinaryTreeNode2(5); // 11        BinaryTreeNode2 n6 = new BinaryTreeNode2(6); // 13        BinaryTreeNode2 n7 = new BinaryTreeNode2(7); // 15        BinaryTreeNode2 n8 = new BinaryTreeNode2(8); // 4        BinaryTreeNode2 n9 = new BinaryTreeNode2(9); // 2        BinaryTreeNode2 n10 = new BinaryTreeNode2(10); // 5        BinaryTreeNode2 n11 = new BinaryTreeNode2(11); // 1        BinaryTreeNode2 n12 = new BinaryTreeNode2(12); // 6        BinaryTreeNode2 n13 = new BinaryTreeNode2(13); // 3        BinaryTreeNode2 n14 = new BinaryTreeNode2(14); // 7        BinaryTreeNode2 n15 = new BinaryTreeNode2(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));    }}
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