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Breadth-first Search - Easy Level - Question 1

Breadth-first Search - Easy Level - Question 1


Leetcode 101 Binary Tree Level Order Traversal

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

Constraints:

The number of nodes in the tree is in the range [0, 2000].

-1000 <= Node.val <= 1000


Analysis:

This question is pretty-straightforward: starting with the root, then layer-by-layer goes down. 

1. If use a queue, need to know the number of nodes in the current nodes;

2. for each node at one layer, record the node value, then save its child leaves if exist.


See the code below:


/**
 * Definition for binary tree
 * 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) {
        vector<vector<int>> res;
        if(!root) return res;
        queue<TreeNode *> q;
        q.push(root);
        while(!q.empty()) {
            vector<int> level;
            int k = q.size();
            for(int i=0; i<k; ++i) {
                auto t = q.front();
                q.pop();
                level.push_back(t->val);
                if(t->left) q.push(t->left);
                if(t->right) q.push(t->right);
            }
            res.push_back(level);
        }
        return res;
    }
};

 


Upper Layer

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