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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,#,#,15,7},
3
/ 9 20
/ 15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
思路：dfs。这里要注意的是，vector数组的size函数返回值是unsigned int类型。因此，语句res.size() - 1 < 0 就会因为溢出变成最大值从而变成false。
1 class Solution { 2 public: 3 void help(vector<vector<int> >& res, TreeNode* root, int depth) 4 { 5 if (root == NULL) return; 6 if (res.size() < depth + 1) 7 { 8 vector<int> tem; 9 res.push_back(tem); 10 } 11 res[depth].push_back(root->val); 12 help(res, root->left, depth + 1); 13 help(res, root->right, depth + 1); 14 } 15 vector<vector<int>> levelOrder(TreeNode* root) { 16 vector<vector<int> > res; 17 help(res, root, 0); 18 return res; 19 } 20 };
Binary Tree Level Order Traversal -- LeetCode
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原文地址：http://www.cnblogs.com/fenshen371/p/5162040.html
