leetcode — populating-next-right-pointers-in-each-node

/**
 *
 * Source : https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node/
 * Source : https://oj.leetcode.com/problems/populating-next-right-pointers-in-each-node-ii/
 *
 *
 *
 * Given a binary tree
 *
 *     struct TreeLinkNode {
 *       TreeLinkNode *left;
 *       TreeLinkNode *right;
 *       TreeLinkNode *next;
 *     }
 *
 * Populate each next pointer to point to its next right node.
 * If there is no next right node, the next pointer should be set to NULL.
 *
 * Initially, all next pointers are set to NULL.
 *
 * Note:
 *
 * You may only use constant extra space.
 * You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
 *
 * For example,
 * Given the following perfect binary tree,
 *
 *          1
 *        /  \
 *       2    3
 *      / \  / \
 *     4  5  6  7
 *
 * After calling your function, the tree should look like:
 *
 *          1 -> NULL
 *        /  \
 *       2 -> 3 -> NULL
 *      / \  / \
 *     4->5->6->7 -> NULL
 *
 * 下面是II，如果是一般二叉树而不是特殊的完全二叉树，求解，因为适合II的解同样可以解决问题I，这里直接解决问题II
 *
 *
 * * Follow up for problem "Populating Next Right Pointers in Each Node".
 * What if the given tree could be any binary tree? Would your previous solution still work?
 *
 * Note:
 * You may only use constant extra space.
 *
 * For example,
 * Given the following binary tree,
 *
 *          1
 *        /  \
 *       2    3
 *      / \    \
 *     4   5    7
 *
 * After calling your function, the tree should look like:
 *
 *          1 -> NULL
 *        /  \
 *       2 -> 3 -> NULL
 *      / \    \
 *     4-> 5 -> 7 -> NULL
 *
 */
public class PopulatingNextRightPointers {

    /**
     *
     * 将二叉树的每一层节点连接成为一个单向链表
     *
     * 假设当前层已经连接好，那么下一层怎么连接？
     * 从当前层的最左节点开始
     *      如果当前根节点root的左右子树都为空，则直接向后移动一个节点
     *
     *      leftMost = root.left != null ? root.left : root.right
     *      cur = leftMost
     *
     *      循环当前一层的所有节点，知道2下一个节点为空，说明当前层循环完
     *      左右子树中至少有一个，：
     *          如果cur==左子树，如果右子树不为空则将cur.next = root.right，且将cur和root向后移，cur = cur.next,root = root.next
     *          如果cur == 右子树，直接将根节点向后移root = root.next
     *          如果cur和左右子树都不相同，则说明cur属于上一个根节点的子节点
     *              如果当前根节点的左右子节点都为空，根节点向后移
     *              否则更新cur，cur = root.left != null ? root.left : root.right,cur = cur.next
     *
     *      对下一层（上面已经连接成链表）进行递归
     *
     * @param root
     * @return
     */
    public TreeLinkedNode populate (TreeLinkedNode root) {
        // 没有子节点，将当前层向后移动一个
        while (root != null && root.leftChild == null && root.rightChild == null) {
            root = root.next;
        }
        if (root == null) {
            return null;
        }
        TreeLinkedNode leftMost = root.leftChild != null ? root.leftChild : root.rightChild;
        TreeLinkedNode cur = leftMost;

        while (root != null) {
            if (cur == root.leftChild) {
                if (root.rightChild != null) {
                    cur.next = root.rightChild;
                    cur = cur.next;
                }
                root = root.next;
            } else if (cur == root.rightChild) {
                root = root.next;
            } else {
                if (root.leftChild == null && root.rightChild == null) {
                    root = root.next;
                } else {
                    cur.next = root.leftChild != null ? root.leftChild : root.rightChild;
                    cur = cur.next;
                }
            }
        }

        populate(leftMost);
        return root;
    }

    public TreeLinkedNode populateByIterator (TreeLinkedNode root) {
        TreeLinkedNode leftMost = root;
        while (leftMost != null) {
            root = leftMost;
            // 没有子节点，将当前层向后移动一个
            while (root != null && root.leftChild == null && root.rightChild == null) {
                root = root.next;
            }
            if (root == null) {
                return null;
            }
            leftMost = root.leftChild != null ? root.leftChild : root.rightChild;
            TreeLinkedNode cur = leftMost;

            while (root != null) {
                if (cur == root.leftChild) {
                    if (root.rightChild != null) {
                        cur.next = root.rightChild;
                        cur = cur.next;
                    }
                    root = root.next;
                } else if (cur == root.rightChild) {
                    root = root.next;
                } else {
                    if (root.leftChild == null && root.rightChild == null) {
                        root = root.next;
                    } else {
                        cur.next = root.leftChild != null ? root.leftChild : root.rightChild;
                        cur = cur.next;
                    }
                }
            }
        }
        return root;
    }

    private class TreeLinkedNode {
        int value;
        TreeLinkedNode next;
        TreeLinkedNode leftChild;
        TreeLinkedNode rightChild;

        public TreeLinkedNode(int value) {
            this.value = value;
        }
    }

    public TreeLinkedNode createTree (char[] treeArr) {
        TreeLinkedNode[] tree = new TreeLinkedNode[treeArr.length];
        for (int i = 0; i < treeArr.length; i++) {
            if (treeArr[i] == '#') {
                tree[i] = null;
                continue;
            }
            tree[i] = new TreeLinkedNode(treeArr[i]-'0');
        }
        int pos = 0;
        for (int i = 0; i < treeArr.length && pos < treeArr.length-1; i++) {
            if (tree[i] != null) {
                tree[i].leftChild = tree[++pos];
                if (pos < treeArr.length-1) {
                    tree[i].rightChild = tree[++pos];
                }
            }
        }
        return tree[0];
    }

    public static void main(String[] args) {
        PopulatingNextRightPointers nextRightPointers = new PopulatingNextRightPointers();
        char[] arr = new char[]{'1','2','3','4','5','#','7'};

        TreeLinkedNode root = nextRightPointers.createTree(arr);
        nextRightPointers.populate(root);

        root = nextRightPointers.createTree(arr);
        nextRightPointers.populateByIterator(root);

    }
}