LeetCode - Populating Next Right Pointers in Each Node

题目:

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

思路:

递归,然后把中间的节点给连起来

package tree;

class TreeLinkNode {
int val;
TreeLinkNode left, right, next;
TreeLinkNode(int x) { val = x; }
} public class PopulatingNextRightPointersInEachNode { public void connect(TreeLinkNode root) {
if (root == null) return;
connect(root.left);
connect(root.right);
TreeLinkNode left = root.left;
TreeLinkNode right = root.right;
while (left != null && right != null) {
left.next = right;
left = left.right;
right = right.left;
}
} }
上一篇:Vue+koa2开发一款全栈小程序(6.个人中心)


下一篇:集成框架 javaweb开发平台ssmy_m(生成代码) java struts2 mybatis spring maven jquery