LeetCode: Populating Next Right Pointers in Each Node 解题报告

Populating Next Right Pointers in Each Node Total
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

SOLUTION 1

我们可以用递归处理左右子树. 这个算法可能会消耗O(h)的栈空间。

 /* 试一下 recursion */
     public void connect(TreeLinkNode root) {
         if (root == null) {
             return;
         }

         rec(root);
     }

     public void rec(TreeLinkNode root) {
         if (root == null) {
             return;
         }

         if (root.left != null) {
             root.left.next = root.right;
         }

         if (root.right != null) {
             root.right.next = root.next.left;
         }

         rec(root.left);
         rec(root.right);
     }

2014.1229 redo:

 /*
         3. A recursion version.
     */
     public void connect3(TreeLinkNode root) {
         if (root == null || root.left == null) {
             return;
         }

         root.left.next = root.right;
         root.right.next = root.next == null ? null: root.next.left;

         connect(root.left);
         connect(root.right);
     }

SOLUTION 2

用层次遍历也可以相当容易解出来，而且这种方法用在下一题一点不用变。

但是 这个解法不能符合题意。题目要求我们使用 constant extra space.

 /*
      * 使用level traversal来做。
      * */
     public void connect1(TreeLinkNode root) {
         if (root == null) {
             return;
         }

         TreeLinkNode dummy = new TreeLinkNode(0);
         Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>();
         q.offer(root);
         q.offer(dummy);

         while (!q.isEmpty()) {
             TreeLinkNode cur = q.poll();
             if (cur == dummy) {
                 if (!q.isEmpty()) {
                     q.offer(dummy);
                 }
                 continue;
             }

             if (q.peek() == dummy) {
                 cur.next = null;
             } else {
                 cur.next = q.peek();
             }

             if (cur.left != null) {
                 q.offer(cur.left);
             }

             if (cur.right != null) {
                 q.offer(cur.right);
             }
         }
     }

 2014.1229 redo:

1.

 /*
         1. Iterator.
     */
     public void connect1(TreeLinkNode root) {
         if (root == null) {
             return;
         }

         Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>();
         q.offer(root);

         while (!q.isEmpty()) {
             int size = q.size();

             for (int i = 0; i < size; i++) {
                 TreeLinkNode cur = q.poll();

                 // ERROR 2: forget to determine if root don't have left and right.
                 if (cur.left == null) {
                     return;
                 }

                 cur.left.next = cur.right;
                 cur.right.next = cur.next == null ? null : cur.next.left;
                 // bug 1: should put the offer inside the for loop
                 q.offer(cur.left);
                 q.offer(cur.right);
             }
         }
     }

2.

 /*
         2. Iterator. More simple version.
     */
     public void connect2(TreeLinkNode root) {
         if (root == null) {
             return;
         }

         Queue<TreeLinkNode> q = new LinkedList<TreeLinkNode>();
         q.offer(root);

         while (!q.isEmpty()) {
             int size = q.size();

             for (int i = 0; i < size; i++) {
                 TreeLinkNode cur = q.poll();

                 // bug 1: should judge the size!
                 cur.next = (i == size - 1) ? null: q.peek();

                 if (cur.left != null) {
                     q.offer(cur.left);
                     q.offer(cur.right);
                 }
             }
         }
     }

SOLUTION 3

把层次遍历修改一下，就是下面的解法了，我们使用2个循环，一个指针P1专门记录每一层的最左边节点，另一个指针P2扫描本层，把下一层的链接上。

下层链接完成后，将P1移动到它的左孩子即可。

这个算法的空间复杂度是O(1). 没有额外的空间。

 /*
         The version that only has O(1) space complexity.
     */
     public void connect(TreeLinkNode root) {
         if (root == null) {
             return;
         }

         Iterator(root);
     }

     public void Iterator(TreeLinkNode root) {
         if (root == null) {
             return;
         }

         TreeLinkNode leftEnd = root;

         while(leftEnd != null) {
             // go through the current level and link the next level.
             TreeLinkNode cur = leftEnd;
             while (cur != null) {
                 if (cur.left == null) {
                     break;
                 }

                 cur.left.next = cur.right;
                 // 一定要记得判null.
                 cur.right.next = cur.next == null ? null: cur.next.left;

                 cur = cur.next;
             }

             // get to the next level.
             leftEnd = leftEnd.left;
         }
     }

 2014.1229 redo:

 /*
         4. Another constant iterator version.
     */
     public void connect(TreeLinkNode root) {
         if (root == null) {
             return;
         }

         TreeLinkNode leftEnd = root;
         while (leftEnd != null && leftEnd.left != null) {
             TreeLinkNode cur = leftEnd;
             while (cur != null) {
                 cur.left.next = cur.right;
                 cur.right.next = cur.next == null ? null: cur.next.left;

                 cur = cur.next;
             }

             leftEnd = leftEnd.left;
         }
     }

GITHUB:

https://github.com/yuzhangcmu/LeetCode_algorithm/blob/master/tree/Connect_2014_1229.java