Leetcode-117: Populating Next Right Pointers in Each Node 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

这道题和前面一题Populating Next Right Pointers in Each Node一样，唯一的不同在于，给出的不是一颗完全二叉树，不过解题思路依然是一样的，使用层序遍历。不过在遍历的时候需要考虑到的几种情况：1）h.left != null。此时，h.left.next的指向有四种情况：h.right, k.left, k.right, null，需要分析前三个节点是否为null；2）h.right != null。该节点的指向有三种情况：h.next.left, k.right, null。k为与h在同一层且在h之后，子节点至少有一个不为null的节点。

/** * Definition for binary tree with next pointer. * public class TreeLinkNode { *     int val; *     TreeLinkNode left, right, next; *     TreeLinkNode(int x) { val = x; } * } */public class Solution {    public void connect(TreeLinkNode root) {        if (root == null) return;        TreeLinkNode head = root, cur = null;        while (head != null) {            cur = head;            boolean nextLevel = false;            while (cur != null) {                TreeLinkNode childNext = null;                if (!nextLevel && (cur.left != null || cur.right != null)) {                    nextLevel = true;                    head = cur.left == null ? cur.right : cur.left;                }                                 TreeLinkNode nextParent = cur.next;                while (nextParent != null && nextParent.next != null && (nextParent.left == null && nextParent.right == null)){                    nextParent = nextParent.next;                }                                if (nextParent != null) {                    childNext = nextParent.left == null ? nextParent.right : nextParent.left;                }                                if (cur.right != null) {                    if (cur.left != null)                        cur.left.next = cur.right;                                        cur.right.next = childNext;                }                else if (cur.left != null)                    cur.left.next = childNext;                                cur = nextParent;            }            if (!nextLevel)                break;        }    }}