Given a binary tree

    struct TreeLinkNode {
      TreeLinkNode *left;
      TreeLinkNode *right;
      TreeLinkNode *next;
    }

Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.

Initially, all next pointers are set to NULL.

Note:

• You may only use constant extra space.
• You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).

For example,
Given the following perfect binary tree,

         1
       /  \
      2    3
     / \  / \
    4  5  6  7

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

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Show Similar Problems

/**
 * 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 * first, * prev;
        first = prev = nullptr;
        while(root != nullptr){ // move down
            first = root->left;
            while(root != nullptr){ // move right
                if(root->left){ // not the leaf node
                    if(prev){
                        prev->next = root->left;
                    }
                    root->left->next = root->right;
                    prev = root->right;
                    root = root->next;
                }else{
                    break;
                }
                
            }
            root = first;
            prev = nullptr;
        }
    }
};

7.31.2015 updated

o(1) space complexity algorithm.

reference: http://www.cnblogs.com/felixfang/p/3647898.html

// Space complexity: o(1)
// Make use of the 'next' pointer to implement a level order traversal algorithm.
//
// When we visit a node p, if it's not a leaf node, set p->left->next = p->right which links its two children.
// Besides, we maintain a preChild pointer to store the pre-visited right child, link it to the left child of next node in the upper level.
//
/**
 * 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 = root;
        TreeLinkNode * preChild;
        while(root != NULL){
            preChild = NULL;
            //finish, leaf node
            if(root->left == NULL){
                break;
            }
            p = root;
            while(p != NULL){
                p->left->next = p->right;
                if(preChild != NULL){
                    preChild->next = p->left;
                }
                preChild = p->right;
                p = p->next;
            }
            root = root->left;
        }
    }
};

//Space Complexity: o(log(n)).
//This algorithm uses path[] array to store current visit path in the tree, which is equivalent to level order traversal. 
//It compresses space complexity comparing to the traditional 'queue' algorithm. 
//To be improved. 
//o(1) is required.
/**
 * 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 * pre = NULL;
        TreeLinkNode * p = NULL;
        bool isNewLevel;
        int path[100]; // Maximum 100 levels, if path[0] == 0, carry one to next digit. 0 means left child, 1 means right child.
        int depth = 0; //current depth
        for(int i = 0; i < 100; i++){
            path[i] = -1;
        }
        p = next(root, path, depth, isNewLevel);
        while(p != NULL){
            if(pre == NULL || isNewLevel == true){
                pre = p;
            }
            else{
                pre->next = p;
                pre = p;
            }
            p = next(root, path, depth, isNewLevel);
        }
    }
    TreeLinkNode * next(TreeLinkNode*root, int path[], int& depth, bool& isNewLevel){
        TreeLinkNode * p = root;
        isNewLevel = false;
        if(root == NULL){
            return NULL;
        }
        //base condition
        if(depth == 0){
            depth = 1;
            path[1] = 0;
            isNewLevel = true;
            return root->left;
        }
        path[depth]++;
        int i = depth;
        while(path[i] == 2){
            path[i - 1]++;
            path[i] = 0;
            i--;
        }
        //carry digit
        if(path[0] == 0){
            path[0] = -1;
            depth++;
            path[depth] = 0;
            isNewLevel = true;
        }
        for(int i = 1; i <= depth; i++){
            p = path[i] == 0?p->left:p->right;
        }
        return p;
    }
};