leetCode 116.Populating Next Right Pointers in Each Node (为节点填充右指针) 解题思路和方法
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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:为二叉树的每个节点均添加右指针,指向下一个水平序节点。其思路是根据上一层的右指针来求得下一个水平序节点。
具体代码如下:
/** * 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) { dfs(root); } /** * dfs,通过next来求解,假设上一层next指针均设置好, * 则下一层的next指针也可依次设置 * 因为是完美二叉树,所有的底层左子树均存在,不存在的说明结束 */ private void dfs(TreeLinkNode root){ if(root == null || root.left == null){ return; } TreeLinkNode p = root; while(root != null){ root.left.next = root.right; if(root.next != null){ root.right.next = root.next.left; } root = root.next; } dfs(p.left); }}
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