(二叉树 BFS) 102. Binary Tree Level Order Traversal

Posted 2021-02-16 weixu-liu

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

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]
]

-----------------------------------------------------------------------------
就是遍历二叉树的的每一层，并且把每一层的数加入到一个数组，返回这个二维数组。

C++代码：

/**
 * 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:
    vector<vector<int>> levelOrder(TreeNode* root) {
        queue<TreeNode*> q;
        vector<int> vec;
        vector<vector<int> > vecs;
        if(!root){
            return vecs;   //这个vecs指的是空数组。
        }
        q.push(root);
        while(!q.empty()){
            vec.clear();
            for(int i = q.size();i>0;i--){
                auto t = q.front();
                q.pop();
                vec.push_back(t->val);
                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
            }
            vecs.push_back(vec);
        }
        return vecs;
    }
};