Leetcode 102. Binary Tree Level Order Traversal

文章作者:Tyan
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1. Description

Binary Tree Level Order Traversal

2. Solution

  • Version 1
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/**
* 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;
if(!root) {
return result;
}
queue<TreeNode*> q;
q.push(root);
traverseLevel(q, result);
return result;
}

private:
void traverseLevel(queue<TreeNode*>& q, vector<vector<int>>& result) {
vector<int> values;
int size = q.size();
while(size) {
TreeNode* current = q.front();
q.pop();
if(current) {
values.push_back(current->val);
if(current->left) {
q.push(current->left);
}
if(current->right) {
q.push(current->right);
}
}
size--;
}
result.push_back(values);
if(!q.empty()) {
traverseLevel(q, result);
}
}
};
  • Version 2
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/**
* 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;
if(!root) {
return result;
}
queue<TreeNode*> q1;
queue<TreeNode*> q2;
q1.push(root);
vector<int> values;
while(!q1.empty()) {
TreeNode* current = q1.front();
q1.pop();
if(current) {
values.push_back(current->val);
if(current->left) {
q2.push(current->left);
}
if(current->right) {
q2.push(current->right);
}
}
if(q1.empty()) {
result.push_back(values);
values.clear();
q1 = q2;
q2 = {};
}
}
return result;
}
};

Reference

  1. https://leetcode.com/problems/binary-tree-level-order-traversal/description/
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