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).

Example:

    3
   / \
  9  20
    /  \
   15   7

return its level order traversal as:

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

Solution

Approach 1: BFS.

Approach 2: DFS.

Code (Python)

Approach 1:

from collections import deque

class Solution:
    def levelOrder(self, root):
        """
        :type root: TreeNode
        :rtype: List[List[int]]
        """
        if not root:
            return []
        result = []
        queue = deque([[root]])
        while queue:
            level = queue.popleft()
            level_result = []
            next_level_nodes = []
            for node in level:
                level_result.append(node.val)
                if node.left:
                    next_level_nodes.append(node.left)
                if node.right:
                    next_level_nodes.append(node.right)
            result.append(level_result)
            if next_level_nodes:
                queue.append(next_level_nodes)
        return result

Code (C++)

Approach 1:

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

Approach 2:

// DFS.
class Solution {
private:
    vector<vector<int>> res;
    void doLevelOrder(TreeNode *root, int level) {
        if (root == NULL)
            return;
        if (res.size() < level)
            res.push_back(vector<int>());
        res[level-1].push_back(root->val);
        doLevelOrder(root->left, level + 1);
        doLevelOrder(root->right, level + 1);
    }
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        doLevelOrder(root, 1);
        return res;
    }
};