Tree

117. Populating Next Right Pointers in Each Node II

1. Question

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

2. Implementation

(1) BFS

public class Solution {
    public void connect(TreeLinkNode root) {
        if (root == null) {
            return;
        }

        Queue<TreeLinkNode> queue = new LinkedList<>();
        queue.add(root);

        while (!queue.isEmpty()) {
            int size = queue.size();
            for (int i = 0; i < size; i++) {
                TreeLinkNode curNode = queue.remove();

                if (i == size - 1) {
                    curNode.next = null;
                }
                else {
                   curNode.next = queue.peek(); 
                }

                if (curNode.left != null) {
                    queue.add(curNode.left);
                }

                if (curNode.right != null) {
                    queue.add(curNode.right);
                }
            }
        }
    }
}

(2) Iteration

public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode preNode = null, curNode = root, leftMostNode = null;

        while (curNode != null) {
            while (curNode != null) {

                if (curNode.left != null) {
                    if (preNode != null) {
                        preNode.next = curNode.left;
                    }
                    else {
                        leftMostNode = curNode.left;
                    }
                    preNode = curNode.left;
                }

                if (curNode.right != null) {
                    if (preNode != null) {
                        preNode.next = curNode.right;
                    }
                    else {
                        leftMostNode = curNode.right;
                    }
                    preNode = curNode.right;
                }

                curNode = curNode.next;
            }
            curNode = leftMostNode;
            preNode = null;
            leftMostNode = null;
        }
    }
}

3. Time & Space Complexity

BFS: 时间复杂度: O(n), 空间复杂度: O(w), w是树中拥有最多node的一层中的node个数

Iteration: 时间复杂度: O(n), 空间复杂度: O(1)