leetCode 116.Populating Next Right Pointers in Each Node (为节点填充右指针) 解题思路和方法
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:为二叉树的每一个节点均加入右指针。指向下一个水平序节点。其思路是依据上一层的右指针来求得下一个水平序节点。
详细代码例如以下:
/**
* 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,如果上一层next指针均设置好,
* 则下一层的next指针也可依次设置
* 由于是完美二叉树,全部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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