Lintcode_69 _ hierarchical traversal of a binary tree, lintcode_69 Binary Tree

Lintcode_69 _ hierarchical traversal of a binary tree, lintcode_69 Binary Tree

Layered traversal of Binary Trees

 

• Description
• Notes
• Data
• Evaluation

Returns the hierarchical traversal of a binary tree's node values (layer-by-layer access from left to right)

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Give a binary tree{3,9,20,#,#,15,7}:

  3 / \9  20  /  \ 15   7

Returns the result of its layered traversal:

[  [3],  [9,20],  [15,7]]

/** * Definition of TreeNode: * class TreeNode { * public: *     int val; *     TreeNode *left, *right; *     TreeNode(int val) { *         this->val = val; *         this->left = this->right = NULL; *     } * } */class Solution {public:    /*     * @param root: A Tree     * @return: Level order a list of lists of integer     */    vector<vector<int>> levelOrder(TreeNode * root) {        // write your code here        queue<TreeNode*> q;        vector<vector<int>> ans;        if(root==NULL)            return ans;        int len;        q.push(root);        while(!q.empty())        {            len=q.size();            vector<int> res;            while(len--)            {               TreeNode *p=q.front();               q.pop();               res.push_back(p->val);               if(p->left)                    q.push(p->left);                if(p->right)                    q.push(p->right);                            }            ans.push_back(res);        }        return ans;    }};