Question:

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.
  • Recursive approach is fine, implicit stack space does not count as extra space for this problem.
  • 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  6  7

After calling your function, the tree should look like:

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

Thinking:

  • Method:
    • 当看到需要将所有同一层上的元素连接在一起，我们就会想到使用BFS遍历树。
/**
 * 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) {
        LinkedList<TreeLinkNode> list = new LinkedList<>();
        if(root == null) return;
        list.add(root);
        while(!list.isEmpty()){
            int size = list.size();
            TreeLinkNode node = list.poll();
            if(node.left != null) list.add(node.left);
            if(node.right != null) list.add(node.right);
            for(int i = 0; i < size - 1; i++){
                TreeLinkNode temp = list.poll();
                if(temp.left != null) list.add(temp.left);
                if(temp.right != null) list.add(temp.right);
                node.next = temp;
                node = temp;
            }
            node.next = null;
        }
    }
}

二刷

简化了代码，不用再将root结点单独分析。

/**
 * 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;
        LinkedList<TreeLinkNode> q = new LinkedList<>();
        q.offer(root);
        while(!q.isEmpty()){
            int size = q.size();
            TreeLinkNode pre = null;
            TreeLinkNode next = null;
            for(int i = 0; i < size; i++){
                pre = q.poll();
                next = i == size - 1 ? null: q.peek();
                pre.next = next;
                if(pre.left != null) q.add(pre.left);
                if(pre.right != null) q.add(pre.right);
            }
        }
    }
}