LEETCODE: Populating Next Right Pointers in Each Node II

最新推荐文章于 2025-12-07 21:20:15 发布

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#LEETCODE #二叉树 #算法 #遍历

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本文讨论了如何在不使用额外空间的情况下，为任意给定的二叉树节点添加相邻指针，以实现节点之间的连接。通过队列结构遍历树的层次，确保每个节点正确地指向其下一个节点。

Follow up for problem "Populating Next Right Pointers in Each Node".

What if the given tree could be any binary tree? Would your previous solution still work?

Note:

You may only use constant extra space.

For example,
Given the following binary tree,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \    \
    4-> 5 -> 7 -> NULL

和上一题解法完全相同！

/**
 * 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 connect(TreeLinkNode *root) {
        if(root == NULL) return;
        queue<TreeLinkNode *> level;
        level.push(root);
        while(!level.empty()) {
            queue<TreeLinkNode *> newlevel;
            while(!level.empty()) {
                TreeLinkNode *front = level.front();
                if(front->left != NULL)
                    newlevel.push(front->left);
                if(front->right != NULL)
                    newlevel.push(front->right);
                level.pop();
                if(!level.empty())
                    front->next = level.front();
            }
            level = newlevel;
        }
    }
};