Leetcode: Populating Next Right Pointers in Each Node

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

复杂度

时间 O(N) 空间 O(N)

思路

相当于是Level Order遍历二叉树。通过一个Queue来控制每层的遍历，注意处理该层最后一个节点的特殊情况。此方法同样可解第二题。

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

没想到的方法：因为guarantee了是perfect binary tree

层次递进法

复杂度

时间 O(N) 空间 O(1)

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

        while(cur.left!=null){
            nextLeftmost = cur.left; // save the start of next level
            while(cur!=null){
                cur.left.next=cur.right;
                cur.right.next = cur.next==null? null : cur.next.left;
                cur=cur.next;
            }
            cur=nextLeftmost;  // point to next level 
        }
    }
}