Populating Next Right Pointers in Each Node

最新推荐文章于 2025-12-06 07:43:37 发布

转载最新推荐文章于 2025-12-06 07:43:37 发布 · 433 阅读

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#leetcode

本文介绍了一种在完美二叉树中填充每个节点的next指针的方法，使得每个节点指向其右侧相邻节点。文章提供了一个C++实现方案，通过递归地连接兄弟节点和堂兄弟节点，并将最右侧节点的next指针设置为NULL。

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

/**

* Definition for binary tree with next pointer.
* struct TreeLinkNode {
* int val;
* TreeLinkNode *left, *right, *next;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
void linkSibling(TreeLinkNode * node){
if (node->left && node->right){
node->left->next = node->right;
linkSibling(node->left);
linkSibling(node->right);
}
}

void linkCousin(TreeLinkNode * node){
if (node->right && node->next){
node->right->next = node->next->left;
}
if (node->left)
linkCousin(node->left);
if (node->right)
linkCousin(node->right);
}

void setRight2NULL(TreeLinkNode * node){
if (node){
node->next = NULL;
setRight2NULL(node->right);
}
}

void connect(TreeLinkNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (root==NULL)
return;
else if (root->left == NULL && root->right == NULL){
root->next = NULL;
return;
}

// link sibling
linkSibling(root);

// link cousin
linkCousin(root);

// set most right
setRight2NULL(root);
}
};