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[Leetcode] Populating Next Right Pointers in Each Node
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
Solution: recursive.
1 /** 2 * Definition for binary tree with next pointer. 3 * public class TreeLinkNode { 4 * int val; 5 * TreeLinkNode left, right, next; 6 * TreeLinkNode(int x) { val = x; } 7 * } 8 */ 9 public class Solution {10 public void connect(TreeLinkNode root) {11 if(root==null)12 return;13 if(root.left!=null)14 root.left.next=root.right;15 if(root.right!=null)16 root.right.next=root.next==null?null:root.next.left;17 connect(root.left);18 connect(root.right);19 }20 }
[Leetcode] Populating Next Right Pointers in Each Node
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