LeetCode - Binary Tree Level Order Traversal

Input: root = [3, 9, 20, null, null, 15, 7]Output: [[3], [9, 20], [15, 7]]

Input: root = [1]Output: [[1]]

Input: root = []Output: []

- The number of nodes in the tree is in the range [0, 2000]- -1000 <= Node.val <= 1000

void printLevelOrder(node* root){    int h = height(root);    for (int i = 0; i < h; i++)        printCurrentLevel(root, i);}void printLevel(node* root, int level){    if (root == NULL)        return;    if (level == 0)        cout << root->data << " ";    else if (level > 0) {        printLevel(root->left, level - 1);        printLevel(root->right, level - 1);    }}

- initialize 2D array as vector vector<vector<int>> result- initialize size and i- return result if root == null- initialize queue<TreeNode*> q  - push root to queue : q.push(root)- initialize TreeNode* node for iterating on the tree- loop while( !q.empty() ) // queue is not empty  - initialize vector<int> tmp  - set size = q.size()  - loop for i = 0; i < size; i++    - set node = q.front()    - if node->left      - push in queue: q.push(node->left)    - if node->right      - push in queue: q.push(node->right)    - remove the front node: q.pop()  - push the tmp to result: result.push_back(tmp)- return result

class Solution {public:    vector<vector<int>> levelOrder(TreeNode* root) {        vector<vector<int>> result;        int size, i;        if(root == NULL)            return result;        queue<TreeNode*> q;        q.push(root);        TreeNode* node;        while(!q.empty()){            vector<int> tmp;            size = q.size();            for(i = 0; i < size; i++){                node = q.front();                if(node->left)                    q.push(node->left);                if(node->right)                    q.push(node->right);                q.pop();                tmp.push_back(node->val);            }            result.push_back(tmp);        }        return result;    }};

func levelOrder(root *TreeNode) [][]int {    result := [][]int{}    queue := []*TreeNode{root}    for len(queue) != 0 {        tmp := []int{}        size := len(queue)        for i := 0; i < size; i++ {            if queue[0] != nil {                tmp = append(tmp, queue[0].Val)                queue = append(queue, queue[0].Left)                queue = append(queue, queue[0].Right)            }            queue = queue[1:]        }        result = append(result, tmp)    }    return result[:len(result)-1]}

var levelOrder = function(root) {    let result = [];    let queue = [];    if(root)        queue.push(root);    while(queue.length > 0) {        tmp = [];        let len = queue.length;        for (let i = 0; i< len; i++) {            let node = queue.shift();            tmp.push(node.val);            if(node.left) {                queue.push(node.left);            }            if(node.right) {                queue.push(node.right);            }        }        result.push(tmp);    }    return result;};

Input: root = [3, 9, 20, null, null, 15, 7]Step 1: vector<vector<int>> result;        int size, i;Step 2: root == null        [3, 9..] == null        falseStep 3: queue<TreeNode*> q;        q.push(root);        q = [3]Step 4: loop !q.empty()        q = [3]        q.empty() = false        !false = true        vector<int> tmp        size = q.size()             = 1        for(i = 0; i < 1; i++)          - 0 < 1          - true          node = q.front()          node = 3          if node->left            - node->left = 9            - q.push(node->left)            - q = [3, 9]          if node->right            - node->right = 20            - q.push(node->right)            - q = [3, 9, 20]          q.pop()          q = [9, 20]          tmp.push_back(node->val)          tmp.push_back(3)          i++          i = 1        for(i < 1)        1 < 1        false        result.push_back(tmp)        result = [[3]]Step 5: loop !q.empty()        q = [9, 20]        q.empty() = false        !false = true        vector<int> tmp        size = q.size()             = 2        for(i = 0; i < 2; i++)          - 0 < 2          - true          node = q.front()          node = 9          if node->left            - node->left = nil            - false          if node->right            - node->right = nil            - false          q.pop()          q = [20]          tmp.push_back(node->val)          tmp.push_back(9)          i++          i = 1        for(i < 2)          - 1 < 2          - true          node = q.front()          node = 20          if node->left            - node->left = 15            - q.push(node->left)            - q = [20, 15]          if node->right            - node->left = 7            - q.push(node->right)            - q = [20, 15, 7]          q.pop()          q = [15, 7]          tmp.push_back(node->val)          tmp.push_back(20)          tmp = [9, 20]          i++          i = 2        for(i < 2)          - 2 < 2          - false        result.push_back(tmp)        result = [[3], [9, 20]]Step 6: loop !q.empty()        q = [15, 7]        q.empty() = false        !false = true        vector<int> tmp        size = q.size()             = 2        for(i = 0; i < 2; i++)          - 0 < 2          - true          node = q.front()          node = 15          if node->left            - node->left = nil            - false          if node->right            - node->right = nil            - false          q.pop()          q = [7]          tmp.push_back(node->val)          tmp.push_back(15)          i++          i = 1        for(i < 2)          - 1 < 2          - true          node = q.front()          node = 7          if node->left            - node->left = nil            - false          if node->right            - node->right = nil            - false          q.pop()          q = []          tmp.push_back(node->val)          tmp.push_back(7)          tmp = [15, 7]          i++          i = 2        for(i < 2)          - 2 < 2          - false        result.push_back(tmp)        result = [[3], [9, 20], [15, 7]]Step 7: loop !q.empty()        q = []        q.empty() = true        !true = falseStep 8: return resultSo we return the result as [[3], [9, 20], [15, 7]].