leetcode解题: Populating next right pointers in each node II (117)

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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,
1
/ \
2 3
/ \ \
4 5 7
After calling your function, the tree should look like:
1 -> NULL
/ \
2 -> 3 -> NULL
/ \ \
4-> 5 -> 7 -> NULL

解法1： BFS， O（N） Time + O（N） Space

运用此前perfect tree的解法，此题任然有效。
C++

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void connect(TreeLinkNode *root) {
    if (!root) {
        return;
    }
    queue<TreeLinkNode*> q;
    q.push(root);
    while(!q.empty()) {
        int number = q.size();
        TreeLinkNode* prev = NULL;
        for (int i = 0; i < number; ++i) {
            TreeLinkNode* cur = q.front();
            q.pop();
            cur->next = prev;
            prev = cur;
            if (cur->right) {
                q.push(cur->right);
            }
            if (cur->left) {
                q.push(cur->left);
            }
        }
    }
    return;
}

Java

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解法2： Iterative， O（N） Time + O（1） Space

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public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode dummy = new TreeLinkNode(0);   // point to the next level's first node
        TreeLinkNode prev = dummy;
        while (root != null) {
            if (root.left != null) {
                prev.next = root.left;
                prev = prev.next;
            }
            if (root.right != null) {
                prev.next = root.right;
                prev = prev.next;
            }
            root = root.next;
            if (root == null) {
                // end of current level, move to the next level
                root = dummy.next;
                prev = dummy;
                dummy.next = null;
            }
        }
    }
}