LintCode (69) Binary Tree Level Order Traversal

Medium Binary Tree Level Order Traversal

32%

Accepted

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

Have you met this question in a real interview? 

Yes

Example

Given binary tree {3,9,20,#,#,15,7},

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

Challenge

Using only 1 queue to implement it.

btree的level order traversal有个模板。。。就是维护一个Q， 然后用BFS遍历。。。如果熟悉flood fill就知道这个办法了。。。

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/** * Definition of TreeNode: * class TreeNode { * public: * int val; * TreeNode *left, *right; * TreeNode(int val) { * this->val = val; * this->left = this->right = NULL; * } * } */ class Solution { /** * @param root: The root of binary tree. * @return: Level order a list of lists of integer */ public: vector<vector<int>> levelOrder(TreeNode *root) { vector<vector<int>> res; if (!root) return res; queue<TreeNode*> q; q.push(root); while(!q.empty()){ int size=q.size(); vector<int> level; for (int i=0; i<size; i++){ TreeNode* cur=q.front(); q.pop(); if (cur->left) q.push(cur->left); if (cur->right) q.push(cur->right); level.push_back(cur->val); } res.push_back(level); } return res; } };