#69 Binary Tree Level Order Traversal_69 路 binary tree level order traversal-优快云博客

本文链接：https://blog.youkuaiyun.com/haifischxia/article/details/52343482

本文介绍了一种解决二叉树层次遍历问题的方法，并提供了使用广度优先搜索（BFS）实现的具体代码示例。文章还提出了两项挑战：一是只使用一个队列来实现该算法；二是使用深度优先搜索（DFS）算法来完成任务。

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题目描述：

Given a binary tree, return the level order traversal of its nodes' values. (ie, from left to right, level by level).

Have you met this question in a real interview?

Yes

Example

Given binary tree {3,9,20,#,#,15,7},

return its level order traversal as:

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

Challenge

Challenge 1: Using only 1 queue to implement it.

Challenge 2: Use DFS algorithm to do it.

题目思路：

这题比#70更直接，用BFS之后直接输出ans就行。

Mycode（AC = 14ms）：

/**
 * Definition of TreeNode:
 * class TreeNode {
 * public:
 *     int val;
 *     TreeNode *left, *right;
 *     TreeNode(int val) {
 *         this->val = val;
 *         this->left = this->right = NULL;
 *     }
 * }
 */
 
 
class Solution {
    /**
     * @param root: The root of binary tree.
     * @return: Level order a list of lists of integer
     */
public:
    vector<vector<int>> levelOrder(TreeNode *root) {
        // write your code here
        vector<vector<int>> ans;
        if (root == NULL) return ans;
        
        // initialize 2 queues, one to 
        // store tree nodes, one to store
        // level/height of corresponding node
        queue<TreeNode *> helper;
        helper.push(root);
        queue<int> levels;
        levels.push(0);
        
        while (!helper.empty()) {
            TreeNode *node = helper.front();
            helper.pop();
            
            int level = levels.front();
            levels.pop();
            
            // push the node into ans, according
            // to its level
            if (ans.size() <= level) {
                ans.push_back({node->val});
            }
            else {
                ans[level].push_back(node->val);
            }
            
            // push the left/right tree and their
            // corresponding level into queue
            if (node->left) {
                helper.push(node->left);
                levels.push(level + 1);
            }
            if (node->right) {
                helper.push(node->right);
                levels.push(level + 1);
            }
        }
        
        return ans;
    }
};