Binary Tree Level Order Traversal

文章目录

1. Binary Tree Level Order Traversal

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},

return its level order traversal as:

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

思路

队列，BFS

解题

c++版

/**
 * 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>> result;
        if(!root)
            return result;
        vector<int> res;
        queue<TreeNode*> que;
        que.push(root);
        int count=que.size();
        while(!que.empty()){
            TreeNode* tmp=que.front();
            que.pop();
            count--;
            res.push_back(tmp->val);
            if(tmp->left)
                que.push(tmp->left);
            if(tmp->right)
                que.push(tmp->right);
            if(count==0){
                result.push_back(res);
                count=que.size();
                res.clear();
            }
        }
        return result;
    }
};

python版

# Definition for a  binary tree node
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    # @param root, a tree node
    # @return a list of lists of integers
    def levelOrder(self, root):
        if not root:
            return []
        result=[]
        queue=[]
        queue.append(root)
        count=len(queue)
        s=[]
        while len(queue)!=0:
            tmp=queue[0]
            s.append(tmp.val)
            del queue[0]
            count-=1
            if tmp.left:
                queue.append(tmp.left)
            if tmp.right:
                queue.append(tmp.right)
            if count==0:
                result.append(s)
                s=[]
                count=len(queue)
        return result