Populating Next Right Pointers in Each Node

问题描述

 

 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

 

解决思路

层次遍历。

开始节点为根节点，下一层为节点的左子结点。

连接的过程中注意提前判断节点非空。

 

程序

 

public class Solution {
    public void connect(TreeLinkNode root) {
        if (root == null) {
            return ;
        }

        TreeLinkNode curLevel = root;

        while (curLevel != null) {
            TreeLinkNode cur = curLevel;
            while (cur != null) {
                TreeLinkNode next = cur.next;
                if (cur.left != null && cur.right != null) {
                    cur.left.next = cur.right;
                }
                if (cur.right != null && next != null) {
                    cur.right.next = next.left;
                }
                cur = next;
            }
            curLevel = curLevel.left;
        }
    }
}

 

Follow up

 

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

程序

public class Solution {
    public void connect(TreeLinkNode root) {
        if (root == null) {
            return ;
        }
        
        TreeLinkNode curLevel = root;
        
        while (curLevel != null) {
            TreeLinkNode cur = curLevel;
            TreeLinkNode nextLevel = null; // mark the next node
            while (cur != null) {
                TreeLinkNode next = cur.next;
                while (next != null && next.left == null && next.right == null) {
                    next = next.next; // skip node with no children
                }
                if (nextLevel == null) {
                    // mark the first no-empty node as the next node
                    if (cur.left != null) {
                        nextLevel = cur.left;
                    } else if (cur.right != null) {
                        nextLevel = cur.right;
                    }
                }
                if (cur.left != null && cur.right != null) {
                    cur.left.next = cur.right;
                }
                if (next != null) {
                    // connect cross the subtrees
                    if (cur.right != null) {
                        cur.right.next = next.left == null ? next.right : next.left;
                    } else if (cur.left != null) {
                        cur.left.next = next.left == null ? next.right : next.left;
                    }
                }
                cur = next;
            }
            curLevel = nextLevel;
        }
    }
}