【LeetCode 102】Binary Tree Level Order Traversal (C++)

Problem:

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,null,null,15,7],

3
/ 
9
20
/

15
7

return its level order traversal as:

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

Solution:

用一个队列实现，很难用栈实现，因为需要保证同层的有序插入

/**
* Definition for a binary tree node.
* 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>> result;
queue<TreeNode*> q;
if(root==NULL)return result;
q.push(root);
while(!q.empty()){
result.push_back(vector<int>());
int s,deep;
s=q.size();
deep=result.size()-1;
while(s>0){
TreeNode* t=q.front();
q.pop();
result[deep].push_back(t->val);
if(t->left!=NULL)q.push(t->left);
if(t->right!=NULL)q.push(t->right);
s--;
}
}
return result;
}
};

Attention:

        stl数据结构vector和queue的使用

最后

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