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

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It’s similar to the former problem Populating Next Right Pointers in Each Node I. The key to this question is to implement two functions: findNextLevelStartNode and findNextLevelNextNode

/**
 * 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) {
        TreeLinkNode * p, * startNode;
        TreeLinkNode * q;
        while((startNode = findNextLevelStartNode(root)) != NULL){
             p = startNode;
             while((q = findNextLevelNextNode(root, p)) != NULL ){
                    p->next = q;
                    p = q;
             }
             root = startNode;
        }
    } 
    //find the leftmost node in next level
    //if return NULL, traversal is end
    TreeLinkNode * findNextLevelStartNode(TreeLinkNode * root){
        if(root == NULL){
            return NULL;
        }
        if(root->left != NULL) return root->left;
        if(root->right != NULL) return root->right;
        return findNextLevelStartNode(root->next);
    }
    //find next node
    //if return NULL, current level is end
    TreeLinkNode * findNextLevelNextNode(TreeLinkNode *& root, TreeLinkNode * node){
        if(root == NULL){
            return NULL;
        }
        if(node == NULL){
            if(root->left != NULL) return root->left;
            if(root->right != NULL) return root->right;
            root = root->next;
            return findNextLevelNextNode(root, node);
        }
        if(node == root->left){
            if(root->right != NULL) return root->right;
            else{
                root = root->next;
                return findNextLevelNextNode(root, node);
            }
        }
        if(node == root->right){
            root = root->next;
            return findNextLevelNextNode(root, node);
        }
        if(root->left != NULL) return root->left;
        if(root->right != NULL) return root->right;
        root = root->next;
        return findNextLevelNextNode(root, node);
    }
};