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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
广搜+递归
递归两次
connectLR()后的树如图
1 -> NULL / 2 3 -> NULL / \ / 4->5 6->7 -> NULLconnectRL()用于连接右子树到左子树 运行时间:436ms
/** * 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; connectLR(root); connectRL(root); } private void connectLR(TreeLinkNode root){ if(root.left==null) return; root.left.next = root.right; connectLR(root.left); connectLR(root.right); } private void connectRL(TreeLinkNode root){ if(root.left==null) return; if(root.next!=null){ root.right.next = root.next.left; } connectRL(root.left); connectRL(root.right); } }
一次递归完成 运行时间:440ms
/** * 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) { connect(root,null); } private void connect(TreeLinkNode root, TreeLinkNode brother){ if(root == null) return; root.next = brother; connect(root.left,root.right); if(brother!=null){ connect(root.right,brother.left); }else{ connect(root.right,null); } } }
[LeetCode]Populating Next Right Pointers in Each Node
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