LeetCode 102. Binary Tree Level Order Traversal

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]
]

/**
 * 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<int> row;
        vector<vector<int>> v;
        queue<TreeNode *> q;
        if(root == NULL)
            return v;
        q.push(root);
        TreeNode *temp;
        while(!q.empty()) {
            int size = q.size();
            while(size--) {
                temp = q.front();
                q.pop();
                row.push_back(temp->val);
                if(temp->left != NULL) {
                    q.push(temp->left);
                }
                if(temp->right != NULL) {
                    q.push(temp->right);
                }
            }
            v.push_back(row);
            row.clear();
        }
        return v;
    }
};

/**

* 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<int> row;

vector<vector<int>> v;

queue<TreeNode *> q;

if(root == NULL)

return v;

q.push(root);

TreeNode *temp;

while(!q.empty()) {

int size = q.size();

while(size--) {

temp = q.front();

q.pop();

row.push_back(temp->val);

if(temp->left != NULL) {

q.push(temp->left);

}

if(temp->right != NULL) {

q.push(temp->right);

}

v.push_back(row);

row.clear();

}

return v;

}

};

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