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  • 62-Binary Tree Level Order Traversal

    1. Binary Tree Level Order Traversal My Submissions QuestionEditorial Solution
      Total Accepted: 102531 Total Submissions: 312211 Difficulty: Easy
      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},
    3
    /
    9 20
    /
    15 7
    return its level order traversal as:
    [
    [3],
    [9,20],
    [15,7]
    ]

    思路:用队列存储每一层的节点,出队时,将下一层入新队,依次遍历每层
    时间O(n),空间O(n)

    /**
     * Definition for a binary tree node.
     * 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) {
            vector<vector<int>> res;
            if(root==NULL)return res;  //root为空的情况先处理,也可在while条件中加限制
            queue<TreeNode*> levelnode;
            vector<int> row;
            levelnode.push(root);
            while(!levelnode.empty()){
                queue<TreeNode*> pre;
                vector<int> tmp;
                while(!levelnode.empty()){
                    TreeNode * treetmp=levelnode.front();
                    levelnode.pop();
                    tmp.push_back(treetmp->val);
                    if(treetmp->left!=NULL)pre.push(treetmp->left);
                    if(treetmp->right!=NULL)pre.push(treetmp->right);
                }
                res.push_back(tmp);
                levelnode = pre;
            }
            return res;
        }
    };
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  • 原文地址:https://www.cnblogs.com/freeopen/p/5482896.html
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