LeetCode:Binary Tree Level Order Traversal

Binary Tree Level Order Traversal

Total Accepted: 105297 Total Submissions: 318601 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]]

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

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 (M) Binary Tree Zigzag Level Order Traversal (E) Binary Tree Level Order Traversal II (E) Minimum Depth of Binary Tree(M) Binary Tree Vertical Order Traversal

思路：

层次遍历，BFS。

新

java code:

/** * Definition for a binary tree node. * public class TreeNode { *     int val; *     TreeNode left; *     TreeNode right; *     TreeNode(int x) { val = x; } * } */public class Solution {    public List<List<Integer>> levelOrder(TreeNode root) {                Queue<TreeNode> queue = new LinkedList<TreeNode>();        List<List<Integer>> ans = new LinkedList<List<Integer>>();                if(root == null) return ans;                queue.offer(root);        while(!queue.isEmpty()) {                        int size = queue.size();            List<Integer> subAns = new LinkedList<Integer>();            for(int i=0;i<size;i++) {                TreeNode tmp = queue.poll();                subAns.add(tmp.val);                if(tmp.left != null) queue.offer(tmp.left);                if(tmp.right != null) queue.offer(tmp.right);                            }            ans.add(subAns);        }        return ans;    }}

c++ code:

/** * 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>> ret;        if(NULL == root) return ret;        queue<TreeNode *> q[2];        stack<vector<int>> s;        int cur=0;        q[cur].push(root);        while(!q[cur].empty()) {            vector<int> tmp;            while(!q[cur].empty()) {                TreeNode *p = q[cur].front();                tmp.push_back(p->val);                if(p->left) q[cur^1].push(p->left);                if(p->right) q[cur^1].push(p->right);                q[cur].pop();            }            cur^=1;            ret.push_back(tmp);        }        return ret;    }};