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Populating Next Right Pointers in Each Node I, II

Populating Next Right Pointers in Each Node I, II

题目I

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

思路

直接在原完全二叉树上操作,利用某些技巧,具体看代码

/**
 * 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) {
        if(root == NULL) return ;

        TreeLinkNode* temp = root;
        TreeLinkNode* curLne;

        while(temp != NULL) {
            curLne = temp;
            while(curLne != NULL) {
               if(curLne->left != NULL)
               curLne->left->next = curLne->right;

               if(curLne->next != NULL && curLne->right != NULL)
               curLne->right->next = curLne->next->left;

               curLne = curLne->next;
            }

            temp = temp->left;
        }
    }
};

题目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

思路

题目多加了一个要求,就是树可以没有左节点,树已经不是完全二叉树了。

代码多了两个函数,一个函数找到每层开始的节点, 一个函数找到下一个节点

class Solution {
public:
    void connect(TreeLinkNode *root) {
        if(root == NULL) return;
        TreeLinkNode *start;
        TreeLinkNode *current;
        while(root) {
            start = findStart(root);
            current = start;

            while(current) {
                current->next = findNext(root, current);
                current = current->next;
            }

            root = start;
        }
    }

private:
    TreeLinkNode* findNext(TreeLinkNode* root, TreeLinkNode* current) {
        while(root) {
            if(root->left == current) {
                if(root->right != NULL) return root->right;
                root = root->next;
                while(root) {
                    if(root->left != NULL) return root->left;
                    if(root->right != NULL) return root->right;
                    root = root->next;
                }
                return NULL;
            }
            if(root->right == current) {
                root = root->next;
                while(root) {
                    if(root->left != NULL) return root->left;
                    if(root->right != NULL) return root->right;
                    root = root->next;
                }
                return NULL;
            }
            root = root->next;
        }
        return NULL;
    }

    TreeLinkNode* findStart(TreeLinkNode* root) {
        if(root == NULL) return NULL;
        while(root) {
            if(root->left != NULL) return root->left;
            if(root->right != NULL) return root->right;
            root = root->next;
        }
        return NULL;
    }
};

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