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2014
11-18

LeetCode-Populating Next Right Pointers in Each Node II[二叉树]

Populating Next Right Pointers in Each Node 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

标签: Tree Depth-first Search
分析

要处理一个节点,可能需要最右边的兄弟节点,首先想到用广搜。但广搜不是常数空间的,本题要求常数空间。

注意,这题的代码原封不动,也可以解决 Populating Next Right Pointers in Each Node I.

代码1

// LeetCode, Populating Next Right Pointers in Each Node II
// 时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
    void connect(TreeLinkNode *root) {
        if (root == nullptr) return;

        TreeLinkNode dummy(-1);
        for (TreeLinkNode *curr = root, *prev = &dummy; 
                curr; curr = curr->next) {
            if (curr->left != nullptr){
                prev->next = curr->left;
                prev = prev->next;
            }
            if (curr->right != nullptr){
                prev->next = curr->right;
                prev = prev->next;
            }
        }
        connect(dummy.next);
    }
};

代码2

// LeetCode, Populating Next Right Pointers in Each Node II
// 时间复杂度O(n),空间复杂度O(1)
class Solution {
public:
    void connect(TreeLinkNode *root) {
        while (root) {
            TreeLinkNode * next = nullptr; // the first node of next level
            TreeLinkNode * prev = nullptr; // previous node on the same level
            for (; root; root = root->next) {
                if (!next) next = root->left ? root->left : root->right;

                if (root->left) {
                    if (prev) prev->next = root->left;
                    prev = root->left;
                }
                if (root->right) {
                    if (prev) prev->next = root->right;
                    prev = root->right;
                }
            }
            root = next; // turn to next level
        }
    }
};

Java代码:

/**
 * Definition for binary tree with next pointer.
 * public class TreeLinkNode {
 *     int val;
 *     TreeLinkNode left, right, next;
 *     TreeLinkNode(int x) { val = x; }
 * }
 */
public class Solution {
    public void connect(TreeLinkNode root) {
        TreeLinkNode head = root;
        while(head != null){
            TreeLinkNode curNode = head;
            TreeLinkNode tmpNextHead = new TreeLinkNode(0);
            TreeLinkNode pre = tmpNextHead;
            while(curNode != null){
                if(curNode.left != null){
                    pre.next = curNode.left;
                    pre = pre.next;
                }
                if(curNode.right != null){
                    pre.next = curNode.right;
                    pre = pre.next;
                }
                curNode = curNode.next;
            }
            head = tmpNextHead.next;
        }
    }
}

相关题目
Populating Next Right Pointers in Each Node


  1. 约瑟夫也用说这么长……很成熟的一个问题了,分治的方法解起来o(n)就可以了,有兴趣可以看看具体数学的第一章,关于约瑟夫问题推导出了一系列的结论,很漂亮

  2. 这道题这里的解法最坏情况似乎应该是指数的。回溯的时候
    O(n) = O(n-1) + O(n-2) + ….
    O(n-1) = O(n-2) + O(n-3)+ …
    O(n) – O(n-1) = O(n-1)
    O(n) = 2O(n-1)

  3. 这道题目的核心一句话是:取还是不取。
    如果当前取,则index+1作为参数。如果当前不取,则任用index作为参数。

  4. #include <cstdio>

    int main() {
    //answer must be odd
    int n, u, d;
    while(scanf("%d%d%d",&n,&u,&d)==3 && n>0) {
    if(n<=u) { puts("1"); continue; }
    n-=u; u-=d; n+=u-1; n/=u;
    n<<=1, ++n;
    printf("%dn",n);
    }
    return 0;
    }