116. Populating Next Right Pointers in Each Node

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

---

时间复杂度：o(ｎ) 空间复杂度：Ｏ（１），因为记录了next指针，可以在层里查找同层的下一个元素，在做层序遍历时不用用到queue,减少了时间复杂度。

机智，即使是相同的办法，在不同的条件下，可以有优化。比如这题里，想想为啥层序遍历要用ｑｕｅｕｅ，为了找到上一层的下一个元素。如果有了ｎｅｘｔ指针，不用ｑｕｅｕｅ也能找到上一层的下一个元素，得到一个优化解法。 ３０％

/**
 * 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;
        }
        root.next = null;
        TreeLinkNode prevLevelStart = root, prev, cur;
        while (prevLevelStart.left != null) {
            prev = prevLevelStart;
            cur = prevLevelStart.left;
            while (prev != null) {
                if (cur == prev.left) {
                    cur.next = prev.right;
                } else {
                    prev = prev.next;
                    if (prev == null) {
                        cur.next = null;
                    } else {
                        cur.next = prev.left;
                    }
                }
                cur = cur.next;
            }
            prevLevelStart = prevLevelStart.left;
        }
    }
}

 相当与层序遍历，在遍历一层时，把一层里的元素前后连起来。 9% 时间＋空间复杂度: O(n)

/**
 * 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) {
        ArrayDeque<TreeLinkNode> queue = new ArrayDeque<TreeLinkNode>();
        if (root == null) {
            return;
        }
        int size = 0, i = 0;
        TreeLinkNode tmp = null;
        queue.offer(root);
        while (!queue.isEmpty()) {
            size = queue.size();
            i = 0;
            while (i < size) {
                tmp = queue.poll();
                if (tmp.left != null) {
                    queue.offer(tmp.left);
                }
                if (tmp.right != null) {
                    queue.offer(tmp.right);
                }
                if (i != size - 1) {
                    tmp.next = queue.peek();
                } else {
                    tmp.next = null;
                }
                i++;
            }
        }
    }
}