116. Populating Next Right Pointers in Each Node - Medium

标签：start   str   may   int   star   node   ace   开始   nod   

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

用两个指针，start和cur，start标记每一层的起点（最左），cur遍历该层，逐层遍历，start从root开始

time: O(n), space: O(1)

/**
 * 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;
        }
        
        TreeLinkNode start = root, cur = null;
        while(start.left != null) {
            cur = start;
            while(cur != null) {
                cur.left.next = cur.right;
                if(cur.next != null) {
                    cur.right.next = cur.next.left;
                }
                cur = cur.next;
            }
            start = start.left;
        }
    }
}

116. Populating Next Right Pointers in Each Node - Medium

标签：start   str   may   int   star   node   ace   开始   nod   

原文地址：https://www.cnblogs.com/fatttcat/p/10204191.html