[LeetCode] Populating Next Right Pointers in Each Node II

最新推荐文章于 2021-09-24 12:05:40 发布

zjjee

最新推荐文章于 2021-09-24 12:05:40 发布

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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下级链接，以实现树状结构的遍历。通过迭代方法从根节点开始，逐步为每个节点设置指向其子节点的链接。

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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

/**
 * 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;
        TreeLinkNode *p = root -> next;
        while(p){
            if(p -> left){
                p = p -> left;
                break;
            }
            if(p -> right){
                p = p -> right;
                break;
            }
            p = p -> next;
        }
        if(root -> right)
            root ->right -> next = p;
        if(root -> left)
            root -> left -> next = root -> right ? root -> right : p;
        connect(root -> right);
        connect(root -> left);
    }
};