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  • [LeetCode] 117. Populating Next Right Pointers in Each Node II 每个节点的右向指针 II

    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

    116. Populating Next Right Pointers in Each Node 的延续,如果给定的树是任何二叉树,不一定是完全二叉树,就是说不是每个节点都包含有两个子节点。

    解法1:BFS, easy but not constant space, Complexity: time O(N) space O(N) - queue

    解法2: Iteration - use dummy node to keep record of the next level's root to refer pre travel current level by referring to root in the level above,Complexity: time O(N) space O(1)

    Java:BFS

    class Solution {
        public void connect(TreeLinkNode root) {
            if(root == null)return;
            Queue<TreeLinkNode> nodes = new LinkedList<>();
            nodes.offer(root);
            while(!nodes.isEmpty()){
                int size = nodes.size();
                for(int i = 0; i < size; i++){
                    TreeLinkNode cur = nodes.poll();
                    TreeLinkNode n = null;
                    if(i < size - 1){
                        n = nodes.peek();
                    }
                    cur.next = n;
                    if(cur.left != null)nodes.offer(cur.left);
                    if(cur.right != null)nodes.offer(cur.right);
                }
                
            }
        } 
    }
    

    Java: Iteration

    class Solution {
        public void connect(TreeLinkNode root) {
            TreeLinkNode dummy = new TreeLinkNode(0);
            TreeLinkNode pre = dummy;//record next root
            while(root != null){
                if(root.left != null){
                    pre.next = root.left;
                    pre = pre.next;
                }
                if(root.right != null){
                    pre.next = root.right;
                    pre = pre.next;
                }
                root = root.next;//reach end, update new root & reset dummy
                if(root == null){
                    root = dummy.next;
                    pre = dummy;
                    dummy.next = null;
                }
            }
        }
    } 

    Java:

    public void connect(TreeLinkNode root) {
        if(root == null) 
            return;
     
        TreeLinkNode lastHead = root;//prevous level's head 
        TreeLinkNode lastCurrent = null;//previous level's pointer
        TreeLinkNode currentHead = null;//currnet level's head 
        TreeLinkNode current = null;//current level's pointer
     
        while(lastHead!=null){
            lastCurrent = lastHead;
     
            while(lastCurrent!=null){
                //left child is not null
                if(lastCurrent.left!=null)    {
                    if(currentHead == null){
                        currentHead = lastCurrent.left;
                        current = lastCurrent.left;
                    }else{
                        current.next = lastCurrent.left;
                        current = current.next;
                    }
                }
     
                //right child is not null
                if(lastCurrent.right!=null){
                    if(currentHead == null){
                        currentHead = lastCurrent.right;
                        current = lastCurrent.right;
                    }else{
                        current.next = lastCurrent.right;
                        current = current.next;
                    }
                }
     
                lastCurrent = lastCurrent.next;
            }
     
            //update last head
            lastHead = currentHead;
            currentHead = null;
        }
    }

     Python:

    class TreeNode:
        def __init__(self, x):
            self.val = x
            self.left = None
            self.right = None
            self.next = None
        
        def __repr__(self):
            if self is None:
                return "Nil"
            else:
                return "{} -> {}".format(self.val, repr(self.next))
    
    class Solution:
        # @param root, a tree node
        # @return nothing
        def connect(self, root):
            head = root
            while head:
                prev, cur, next_head = None, head, None
                while cur:
                    if next_head is None:
                        if cur.left:
                            next_head = cur.left
                        elif cur.right:
                            next_head = cur.right
                    
                    if cur.left:
                        if prev:
                            prev.next = cur.left
                        prev = cur.left
                        
                    if cur.right:
                        if prev:
                            prev.next = cur.right
                        prev = cur.right
                        
                    cur = cur.next
                head = next_head
    

    C++:

    class Solution {
    public:
        void connect(TreeLinkNode *root) {
            TreeLinkNode *dummy = new TreeLinkNode(0), *t = dummy;
            while (root) {
                if (root->left) {
                    t->next = root->left;
                    t = t->next;
                }
                if (root->right) {
                    t->next = root->right;
                    t = t->next;
                }
                root = root->next;
                if (!root) {
                    t = dummy;
                    root = dummy->next;
                    dummy->next = NULL;
                }
            }
        }
    };
    

      

    类似题目:

    [LeetCode] 116. Populating Next Right Pointers in Each Node 每个节点的右向指针

    All LeetCode Questions List 题目汇总

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  • 原文地址:https://www.cnblogs.com/lightwindy/p/8680870.html
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