卖萌的弱渣

I am stupid, I am hungry.

Populating Next Right Pointers in Each Node

Given a binary tree

1
2
3
4
5

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).

Example

Given the following perfect binary tree,

1
2
3
4
5

     1
   /  \
  2    3
 / \  / \
4  5  6  7

After calling your function, the tree should look like:

1
2
3
4
5

     1 -> NULL
   /  \
  2 -> 3 -> NULL
 / \  / \
4->5->6->7 -> NULL

Solution

  • Java: Better solution
(Populating-Next-Right-Pointers-in-Each-Node.java) download
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32

/**
 * 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 cur = null;
        TreeLinkNode prev = root;
        while(prev.left != null){
            cur = prev;
            // 没一层都连起来
            while(cur!=null){
                cur.left.next = cur.right;
                if(cur.next!=null)
                    cur.right.next = cur.next.left;
                cur = cur.next;
            }

            // 下一层
            prev = prev.left;

        }

    }
}
  • BFS
(Populating-next-right-pointers-in-each-node.py) download
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34

# Definition for binary tree with next pointer.
# class TreeLinkNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None

class Solution(object):
    def connect(self, root):
        """
        :type root: TreeLinkNode
        :rtype: nothing
        """
        if root == None:
            return
        queue = []
        queue.append((root,0))
        oldnode = root
        oldlevel = 0
        while queue:
            item = queue.pop(0)
            cur = item[0]
            level = item[1]
            if cur.left != None:
                queue.append((cur.left,level+1))
            if cur.right != None:
                queue.append((cur.right,level+1))

            if oldnode != cur and oldlevel == level:
                oldnode.next = cur
            oldlevel = level
            oldnode = cur
        return