Leetcode - Populating Next Right Pointers in Each Node II

My code:

/**
 * 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;
        
        Queue q = new LinkedList();
        q.offer(root);
        while (!q.isEmpty()) {
            int levelSize = q.size();
            TreeLinkNode next = null;
            for (int i = 0; i < levelSize; i++) {
                TreeLinkNode temp = q.poll();
                if (i == 0) {
                    temp.next = null;
                    next = temp;
                }
                else {
                    temp.next = next;
                    next = temp;
                }
                if (temp.right != null)
                    q.offer(temp.right);
                if (temp.left != null)
                    q.offer(temp.left);
            }
        }
        
    }
}

My test result:

Paste_Image.png

直接拿上面那道题目的代码，就直接过测试了。。。

**
总结： 右层序遍历树
**

Anyway, Good luck, Richardo!

My code:

/**
 * 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;
        Queue q = new LinkedList();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            TreeLinkNode prev = null;
            for (int i = 0; i < size; i++) {
                TreeLinkNode curr = q.poll();
                if (curr.right != null && curr.left != null) {
                    curr.right.next = prev;
                    curr.left.next = curr.right;
                    prev = curr.left;
                    q.offer(curr.right);
                    q.offer(curr.left);
                }
                else if (curr.right == null && curr.left == null)
                    continue;
                else if (curr.right != null) {
                    curr.right.next = prev;
                    prev = curr.right;
                    q.offer(curr.right);
                }
                else {
                    curr.left.next = prev;
                    prev = curr.left;
                    q.offer(curr.left);
                }
            }
        }
        
    }
}

这次我的代码是过于复杂了点。

Anyway, Good luck, Richardo!

My code:

/**
 * 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;
        }
        
        Queue q = new LinkedList();
        q.offer(root);
        while (!q.isEmpty()) {
            int size = q.size();
            TreeLinkNode pre = null;
            for (int i = 0; i < size; i++) {
                TreeLinkNode curr = q.poll();
                curr.next = pre;
                pre = curr;
                if (curr.right != null) {
                    q.offer(curr.right);
                }
                if (curr.left != null) {
                    q.offer(curr.left);
                }
            }
        }
        
    }
}

以前的做法比较简单，
但是
space: O(n)
time: O(n)

然后看到了
space 优化到 O(1) 的做法，
自己写了下：

My code:

/**
 * 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;
        TreeLinkNode curr = null;
        TreeLinkNode prev = null;
        while (head != null) {
            curr = head;
            head = null;
            prev = null;
            while (curr != null) {
                if (curr.left != null) {
                    if (prev != null) {
                        prev.next = curr.left;
                    }
                    else {
                        head = curr.left;
                    }
                    prev = curr.left;
                }
                if (curr.right != null) {
                    if (prev != null) {
                        prev.next = curr.right;
                    }
                    else {
                        head = curr.right;
                    }
                    prev = curr.right;
                }
                curr = curr.next;
            }
        }
        
    }
}

reference:
https://discuss.leetcode.com/topic/1106/o-1-space-o-n-complexity-iterative-solution/7

其实也是一种BFS的思想

不得不说，我的BFS理解真的很差，解题能力也很差。
赶紧补。

Anyway, Good luck, Richardo! -- 09/06/2016