117. Populating Next Right Pointers in Each Node II-优快云博客

本文链接：https://blog.youkuaiyun.com/BigFatSheep/article/details/83238975

本文介绍了一种迭代算法，用于解决二叉树中同一层节点的连接问题。通过使用常数额外空间，该算法能有效地遍历二叉树并将同一层的节点连接起来，避免了递归带来的额外空间复杂度。

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Description

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.
Example:

Given the following binary tree,

/
2 3
/ \
4 5 7
After calling your function, the tree should look like:

 1 -> NULL

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

Problem URL

Solution

给一棵二叉树，将同一层的二叉树节点连起来。

Solve this problem iteratively. Using cur denotes current node, level denotes that head node of each level, prev denotes previous node in same level. Then start iteration with two while loop. If left != null, judge wether it is the next level’s head with prev, then assign its value to prev. For right child, it is the same. After each level, reset the cur to next level and level and perv are null.

Using this method, would avoid gap in the same level and find right head in next level.

Code

/**
 * 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 cur = root;
        TreeLinkNode level = null;
        TreeLinkNode prev = null;
        while (cur != null){
            while(cur != null){
                if (cur.left != null){
                    if (prev != null){
                        prev.next = cur.left;
                    }
                    else{
                        
                        level = cur.left;
                    }
                    prev = cur.left;
                }
                if (cur.right != null){
                    if (prev != null){
                        prev.next = cur.right;
                    }
                    else{
                        
                        level = cur.right;
                    }
                    prev = cur.right;
                }
                cur = cur.next;
            }
            cur = level;
            level = null;
            prev = null;
        }
    }
}