Leetcode: 102. Sequence traversal of binary tree

102. Sequence traversal of binary tree

source ： Power button （LeetCode）

link : https://leetcode.cn/problems/binary-tree-level-order-traversal/

Give you the root node of the binary tree root , Returns the Sequence traversal . （ That is, layer by layer , Access all nodes from left to right ）.

Example 1：

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

Example 2：

 Input ：root = [1]
 Output ：[[1]]

Example 3：

 Input ：root = []
 Output ：[]

Tips ：

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

solution

• recursive ： A given node And layer , Judge whether the number of layers is the same as the length of the result , If it's the same , Explain the node It should be placed in this layer , Attention is from the 0 Layer start .
• BFS： Sequence traversal , Use a list to maintain the nodes of the current layer . The length of the current layer controls the number of iterations , If there are left and right child nodes , Add it to the list ;

Code implementation

recursive

python Realization

# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        if not root:
            return []
        
        levels = []
        def helper(node, level):
            if len(levels) == level:
                levels.append([])
            levels[level].append(node.val)
            if node.left:
                helper(node.left, level+1)
            
            if node.right:
                helper(node.right, level+1)
        
        helper(root, 0)
        return levels

c++ Realization

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
    vector<vector<int>> res;

public:
    void helper(TreeNode* node, int level) {
    
        if (res.size() == level)
            res.push_back({
    });
        
        res[level].push_back(node->val);
        if (node->left)
            helper(node->left, level+1);
        
        if (node->right)
            helper(node->right, level+1);
    }

    vector<vector<int>> levelOrder(TreeNode* root) {
    
        if (root == nullptr)
            return res;
        
        helper(root, 0);
        return res;
    }

};

Complexity analysis

• Time complexity ： $O(n)$
• Spatial complexity ： $O(n)$

BFS

python Realization

# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, val=0, left=None, right=None):
# self.val = val
# self.left = left
# self.right = right
class Solution:
    def levelOrder(self, root: TreeNode) -> List[List[int]]:
        # bfs
        if not root:
            return []
        
        queue = []
        res = []
        queue.append(root)
        while queue:
            level_res = []
            for i in range(len(queue)):
                node = queue.pop(0)
                level_res.append(node.val)
                if node.left:
                    queue.append(node.left)
                if node.right:
                    queue.append(node.right)
            res.append(level_res)
        return res

c++ Realization

/** * Definition for a binary tree node. * struct TreeNode { * int val; * TreeNode *left; * TreeNode *right; * TreeNode() : val(0), left(nullptr), right(nullptr) {} * TreeNode(int x) : val(x), left(nullptr), right(nullptr) {} * TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {} * }; */
class Solution {
    
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
    
        if (root == nullptr)
            return {
    };
        
        queue<TreeNode*> q;
        vector<vector<int>> res;
        q.push(root);
        while (q.size() != 0) {
    
            vector<int> tmp;
            int level_wide = q.size();
            for(int i=0; i<level_wide; i++) {
    
                TreeNode* node = q.front();
                q.pop();
                tmp.push_back(node->val);
                if(node->left != nullptr)
                    q.push(node->left);
                if (node->right != nullptr)
                    q.push(node->right);
            }
            res.push_back(tmp);
        }
        return res;
    }
};

Complexity analysis

• Time complexity ： $O(n)$ , Enter and leave the team at each point , Therefore, the asymptotic time complexity is O(n)
• Spatial complexity ： $O(n)$, The number of elements in the queue does not exceed n individual , So the asymptotic space complexity is O(n)