zoukankan      html  css  js  c++  java
  • [LeetCode] Populating Next Right Pointers in Each Node II 解题报告


    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
    » Solve this problem

    [解题思路]
    与上一题类似,唯一的不同是每次要先找到一个第一个有效的next链接节点,并且递归的时候要先处理右子树,再处理左子树。


    [Code]
    1:    void connect(TreeLinkNode *root) {  
    2: // Start typing your C/C++ solution below
    3: // DO NOT write int main() function
    4: if(root== NULL) return;
    5: TreeLinkNode* p = root->next;
    6: while(p!=NULL)
    7: {
    8: if(p->left!=NULL)
    9: {
    10: p = p->left;
    11: break;
    12: }
    13: if(p->right!=NULL)
    14: {
    15: p = p->right;
    16: break;
    17: }
    18: p = p->next;
    19: }
    20: if(root->right!= NULL)
    21: {
    22: root->right->next = p;
    23: }
    24: if(root->left !=NULL)
    25: {
    26: root->left->next = root->right? root->right:p;
    27: }
    28: connect(root->right);
    29: connect(root->left);
    30: }


    [Note]
    1. Line 6, while loop, not if
    For example,
     Level 1                      1
                                /              \
     Level 2              2               3
                         /        \                \
     Level 3       4         5                 6
                     /
     Level 4   7

    When processing Level 3, while loop is necessary for finding a valid next.


    Update 03/09/2014 Add a non-recursion version

    1:       void connect(TreeLinkNode *root) {  
    2: TreeLinkNode* cur = root, *next = NULL;
    3: while(cur!=NULL)
    4: {
    5: TreeLinkNode *p = cur, *k= NULL;
    6: while(p!=NULL)
    7: {
    8: TreeLinkNode* sub = getLinkedLeftNode(p);
    9: if(sub != NULL)
    10: {
    11: if(next == NULL)
    12: {
    13: next = sub;
    14: k = sub;
    15: }
    16: else
    17: k->next = sub;
    18: while(k->next !=NULL) // ietrate to the tail
    19: k = k->next;
    20: }
    21: p = p->next;
    22: }
    23: cur = next;
    24: next = NULL;
    25: }
    26: }
    27: TreeLinkNode* getLinkedLeftNode(TreeLinkNode * root)
    28: {
    29: if(root->left != NULL && root->right != NULL)
    30: root->left->next = root->right;
    31: if(root->left != NULL)
    32: return root->left;
    33: if(root->right != NULL)
    34: return root->right;
    35: return NULL;
    36: }


  • 相关阅读:
    Django WSGI响应过程之WSGIHandler
    python多线程
    性能测试概念介绍
    Django惰性加载和LazyObject
    Django中间件分析
    python unittest 快速入门
    jenkins+ant+jmeter测试环境部署
    [Vue warn]: Invalid prop: type check failed for prop "clearable". Expected Boolean, got String with value "true".
    JavaScript通过for循环实现九九乘法表的左下、左上、右上、右下对齐成直角三角形
    找出正确手机号码
  • 原文地址:https://www.cnblogs.com/codingtmd/p/5078974.html
Copyright © 2011-2022 走看看