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  • Binary Tree Level Order Traversal

    题目: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:

    和之前的zigzag读写树的节点数值方法一样,使用堆栈。用循环的方法,保存根节点。

    代码:

    /**
     * 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;
            queue<TreeNode*>q;
            q.push(root);
    
    
            while(!q.empty()){
                int n=q.size();
                vector<int>tem;
                for(int i=0;i<n;i++){
                    TreeNode* temp=q.front();
                    tem.push_back(temp->val);
                    if(temp->left)   q.push(temp->left);
                    if(temp->right)   q.push(temp->right);
                    q.pop();
                }
                res.push_back(tem);
            }
            
            return res;
        }
    };


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  • 原文地址:https://www.cnblogs.com/jsrgfjz/p/8519843.html
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