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116. Populating Next Right Point

116. Populating Next Right Point

作者: 衣介书生 | 来源:发表于2018-04-05 15:44 被阅读10次

    题目分析

    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
    

    这道题目本质上就是完全二叉树的层次遍历的一个变形。

    代码

    import java.util.*;
    /**
     * 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;
            }
            int i = 0;
            int size = (int)Math.pow(2, i);
            Queue<TreeLinkNode> q = new LinkedList<>();
            q.offer(root);
            while(!q.isEmpty()) {
                for(int j = 0; j < size; j++) {
                    TreeLinkNode temp = q.poll();
                    //if(temp.left != null && temp.right != null) {
                    if(temp.left != null) {
                        q.offer(temp.left);
                        q.offer(temp.right);
                    }
                    if(j == size - 1) {
                        temp.next = null;
                    } else {
                        temp.next = q.peek();
                    }
                }
                i += 1;
                size = (int)Math.pow(2, i);
            }
        }
    }
    

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