LeetCode——Binary Tree Level Order Traversal-优快云博客

本文链接：https://blog.youkuaiyun.com/pickless/article/details/11881705

本文介绍了一种使用广度优先搜索（BFS）的方法来实现二叉树的层次遍历，并提供了详细的C++实现代码。通过该方法可以有效地按层次顺序从左到右返回二叉树节点的值。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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},

return its level order traversal as:

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

confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.

» Solve this problem

采用BFS遍历。

/**
 * Definition for binary tree
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function        
        vector<vector<int> > answer;
        if (root != NULL) {
            list<TreeNode *> pre, cur;
            pre.push_back(root);            
            while (!pre.empty()) {
                vector<int> temp;
                for (list<TreeNode *>::iterator iter = pre.begin(); iter != pre.end(); iter++) {
                    temp.push_back((*iter)->val);
                }
                answer.push_back(temp);
                
                cur.clear();
                for (list<TreeNode *>::iterator iter = pre.begin(); iter != pre.end(); iter++) {
                    if ((*iter)->left != NULL) {
                        cur.push_back((*iter)->left);
                    }
                    if ((*iter)->right != NULL) {
                        cur.push_back((*iter)->right);                        
                    }
                }                
                pre = cur;
            }
        }        
        return answer;
    }
};