LeetCode 116. Populating Next Right Pointers in Each Node

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本文详细阐述了如何使用常量额外空间填充完美二叉树节点的next指针，实现节点之间的连接，最终构建出层次化的树结构。

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,

After calling your function, the tree should look like:

         1 -> NULL
       /  \
      2 -> 3 -> NULL
     / \  / \
    4->5->6->7 -> NULL

This one is actually Level-Order Traversal.

#include <iostream>
#include <queue>
using namespace std;

struct TreeLinkNode {
  int val;
  TreeLinkNode *left;
  TreeLinkNode* right;
  TreeLinkNode* next;
  TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
};
/*
             1
          2    3

        changed into

            1->NULL
          2 --> 3 ->NULL
*/
TreeLinkNode* setUpLinkTree() {
  TreeLinkNode* root = new TreeLinkNode(1);
  root->left = new TreeLinkNode(2);
  root->right = new TreeLinkNode(3);
  return root;
}
void connect(TreeLinkNode* root) {
  if(!root) return;
  queue<TreeLinkNode*> nodes;
  queue<TreeLinkNode*> nextLevel;
  nodes.push(root);
  while(!nodes.empty()) {
    TreeLinkNode* tmp = nodes.front();
    nodes.pop();
    if(tmp->left) nextLevel.push(tmp->left);
    if(tmp->right) nextLevel.push(tmp->right);
    while(!nodes.empty()) {
      TreeLinkNode* tmp_next = nodes.front();
      nodes.pop();
      tmp->next = tmp_next;
      tmp = tmp_next;

      if(tmp_next->left) nextLevel.push(tmp_next->left);
      if(tmp_next->right) nextLevel.push(tmp_next->right);
    }
    swap(nodes, nextLevel);
  }
}

void printLinkTree(TreeLinkNode* root) {
  if(!root) return;
  cout << "root value is: " << root->val << " next value is: " << ((root->next == NULL) ? 0 : (root->next)->val) << endl;
  printLinkTree(root->left);
  printLinkTree(root->right);
}

int main(void) {
  TreeLinkNode* root = setUpLinkTree();
  connect(root);
  printLinkTree(root);
}