【LeetCode】 102. Binary Tree Level Order Traversal

Description

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

給一個二元樹，按照階層遍歷回傳它每個節點的值。(從左到右，一層一層)

Example:

For example:
Given binary tree [3,9,20,null,null,15,7],
    3
   / \
  9  20
    /  \
   15   7
return its level order traversal as:
[
  [3],
  [9,20],
  [15,7]
]

Solution

• 使用BFS去跑樹的結構即可。
• 用一個queue去存左子樹和右子樹，同時也去紀錄子樹的level。每次從queue裡面拿新的點來跑，直到queue為空。
• 下方的範例code其實越寫越冗，而且遞迴傳的參數太多，但整體概念和結構是對的。

Code

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    void BFS(vector<vector<int>>& ans,vector<int>& nodes,vector<pair<TreeNode*,int>>& queue,int &level)
    {
        //take new node from queue
        pair<TreeNode*,int> n = queue.front();
        queue.erase(queue.begin());
        if(!n.first)return;
        
        //check level
        if(level==n.second)
            nodes.push_back(n.first->val);
        else
        {
            level = n.second;
            ans.push_back(nodes);
            nodes.clear();
            nodes.push_back(n.first->val);
        }
        //push next node into queue
        queue.push_back(pair<TreeNode*,int>(n.first->left,n.second+1));
        queue.push_back(pair<TreeNode*,int>(n.first->right,n.second+1));
        
        //run until queue is empty
        while(queue.size()!=0)
            BFS(ans,nodes,queue,level);
        
        //check last level is pushed
        if(nodes.size()!=0)
        {
            ans.push_back(nodes);
            nodes.clear();
        }
    }
    
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> ans;
        vector<int> nodes;
        int level=0;
        vector<pair<TreeNode*,int>> queue(1,pair<TreeNode*,int>(root,0));
        BFS(ans,nodes,queue,level);
        return ans;
    }
};

tags: LeetCode C++