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] ]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
其实随便怎么traverse,只需要知道你的level就行了。。。。
/**
* 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;
helper(root,res,0);
return res;
}
void helper(TreeNode* root, vector<vector<int> >& res, int level){
if (root==0)
return;
if (res.size()==level)
res.push_back(vector<int>());
res[level].push_back(root->val);
helper(root->left,res, level+1);
helper(root->right,res,level+1);
}
};
本文介绍了一种解决二叉树层次遍历问题的方法。通过递归的方式记录每一层节点的值,最终得到从上到下、从左到右的层次遍历结果。
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