leetcode 102.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,null,null,15,7],

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  3
 / \
9  20
  /  \
 15   7


return its level order traversal as:
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[
  [3],
  [9,20],
  [15,7]
]

方法一：广度优先遍历

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/**
 * 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) {
        return BFS(root);
    }
private:
    vector<vector<int>> BFS(TreeNode* root) {
        if(!root) return {};
        vector<vector<int>> ans;
        vector<TreeNode*> curr,next;
        curr.push_back(root);
        while(!curr.empty()) {
            ans.push_back({});
            for(TreeNode* node : curr){
                ans.back().push_back(node->val);
                if(node->left)
                    next.push_back(node->left);
                if(node->right)
                    next.push_back(node->right);
            }
            curr.swap(next);
            next.clear();
        }
        return ans;
    }
};
StatusRuntimeLanguage
Accepted4mscpp

方法二:深度优先遍历

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/**
 * 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>> ans;
        DFS(root,0,ans);
        return ans;
    }
private:
    void DFS(TreeNode* root, int depth, vector<vector<int>>& ans) {
        if(!root)
            return;
        while(ans.size()<=depth)
            ans.push_back({});
        DFS(root->left, depth+1, ans);
        DFS(root->right, depth+1, ans);
        ans[depth].push_back(root->val);
    }
};
StatusRuntimeLanguage
Accepted4mscpp