leetcode 117. Populating Next Right Pointers in Each Node II

最新推荐文章于 2023-11-09 13:05:39 发布

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本文介绍了一种在常数额外空间限制下解决“填充每个节点的下一个右侧节点指针”问题的方法，适用于任意二叉树。通过迭代每层的第一个节点和前一个节点，实现了高效的层序遍历。

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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,

After calling your function, the tree should look like:

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

层序遍历就好

public class Solution {
    public void connect(TreeLinkNode root) {
        Queue<TreeLinkNode> queue = new LinkedList<>();
        if(root==null) return;
        queue.offer(root);
        while(!queue.isEmpty()){
            int len = queue.size();
            while(len-->0){
                TreeLinkNode temp = queue.poll();
                if(len==0) temp.next = null;
                else temp.next = queue.peek();
                if(temp.left!=null) queue.offer(temp.left);
                if(temp.right!=null) queue.offer(temp.right);
            }
        }
    }
}

忘了只能用常数空间。。。

参考别人的做法，保存每层的第一个和前一个

public class Solution {
    
    //based on level order traversal
    public void connect(TreeLinkNode root) {

        TreeLinkNode head = null; //head of the next level
        TreeLinkNode prev = null; //the leading node on the next level
        TreeLinkNode cur = root;  //current node of current level

        while (cur != null) {
            
            while (cur != null) { //iterate on the current level
                //left child
                if (cur.left != null) {
                    if (prev != null) {
                        prev.next = cur.left;
                    } else {
                        head = cur.left;
                    }
                    prev = cur.left;
                }
                //right child
                if (cur.right != null) {
                    if (prev != null) {
                        prev.next = cur.right;
                    } else {
                        head = cur.right;
                    }
                    prev = cur.right;
                }
                //move to next node
                cur = cur.next;
            }
            
            //move to next level
            cur = head;
            head = null;
            prev = null;
        }
        
    }
}