LeetCode 117 Populating Next Right Pointers in Each Node II

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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
Solution C++ 


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/**
 * 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* lnode = root;
        while (lnode) {
            TreeLinkNode* node = lnode;
            TreeLinkNode* prev = NULL;
            while (node) {
                if (node->left && node->right) {
                    if (prev) prev->next = node->left;
                    node->left->next = node->right;
                    prev = node->right;
                } else if (node->left) {
                    if (prev) prev->next = node->left;
                    prev = node->left;
                } else if (node->right) {
                    if (prev) prev->next = node->right;
                    prev = node->right;
                }
                node = node->next;
            }
            while (lnode && (lnode->left == NULL) && (lnode->right == NULL)) {
                lnode = lnode->next;
            }
            if (lnode) lnode = lnode->left ? lnode->left: lnode->right;
        }
    }
};
Solution Python 


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# Definition for binary tree with next pointer.
# class TreeLinkNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None
#         self.next = None

class Solution:
    # @param root, a tree link node
    # @return nothing
    def connect(self, root):
        if not root: return
        lnode = root
        while lnode:
            prev = None
            node = lnode
            while node:
                if node.left and node.right:
                    if prev:
                        prev.next = node.left
                    node.left.next = node.right
                    prev = node.right
                elif node.left:
                    if prev:
                        prev.next = node.left
                    prev = node.left
                elif node.right:
                    if prev:
                        prev.next = node.right
                    prev = node.right
                node = node.next
            while lnode and lnode.left is None and lnode.right is None:
                lnode = lnode.next
            if lnode: lnode = lnode.left if (lnode.left) else lnode.right

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