前言
今天上午从9点半开始就做了这么一道题目，还是挺有意思的，这里记录一下题目
Follow up for problem "Populating Next Right Pointers in Each Node".What if the given tree could be any binary tree Would your previous solution still work
Note:
You may only use constant extra space.
For example,
Given the following binary tree,
After calling your function, the tree should look like:
vc/D98/UtcTH+LHwo6zH+LHw1NrT2qO6w7+49r3ateO1xG5leHS92rXjv8nS1NTauuHP8snP1dK1vbTm1Nq1xMTHuPa92rXjo6zO3sLbyse4uL3ateNuZXh0vdq147XE1/O78tXf09K6otfTo6zT1rvy1d/Kx7i4vdq1425leHS1xG5leHS92rXjtcTX87vy1d/T0r3ateOho6GjoaMKPGJyPgoK0vK0y6Os1eK1wMzixL/K19KqysfV0rW909K6otfTtcS12tK7uPbT0NCntcRuZXh0wbS9073ateOjrMi7uvPU2bSmwO3X87qi19Oho8i7uvPSwLTOtd256bSmwO3T0rqi19OjrNfzuqLX0wo8YnI+Cgo8YnI+Cgo8aDI+tPrC6zwvaDI+CjxwcmUgY2xhc3M9"brush:java;">/** * 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) { // Note: The Solution object is instantiated only once and is reused by each test case. if (root == null) { return; } TreeLinkNode p = root.next; while (p != null) { if (p.left != null) { p = p.left; break; } if (p.right != null) { p = p.right; break; } p = p.next; } if (root.right != null) { root.right.next = p; } if (root.left != null) { root.left.next = root.right == null p : root.right; } connect(root.right); connect(root.left); } }
记录
凭借AC这道题目，在LeetCode OJ上ac数目达到50了，截图纪念一下