LeetCode 102. Binary Tree Level Order Traversal 和107

最新推荐文章于 2022-02-20 23:20:19 发布

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最新推荐文章于 2022-02-20 23:20:19 发布

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LeetCode102. Binary Tree Level Order Traversal

题目链接：https://leetcode.com/problems/binary-tree-level-order-traversal/?tab=Description

题目描述：

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],

return its level order traversal as:

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

思路一：用深度优先搜索（DFS），遍历的同时记录深度，将深度相同的插入到结果中，用时6ms

/**
 * 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;
        DFS(root,result,-1);
        return result;
        }
    void DFS(TreeNode* curNode,vector<vector<int>>& result,int curLevel)
    {
        if(curNode==NULL)
            return;
        curLevel++;
        if(result.size()<curLevel+1)
        {
            vector<int> tmp;
            tmp.push_back(curNode->val);
            result.push_back(tmp);
        }
        else
            result[curLevel].push_back(curNode->val);
        DFS(curNode->left,result,curLevel);
        DFS(curNode->right,result,curLevel);
    }
};

思路二：用广度优先遍历（BFS），记录每一层的size，将每一层插入到结果中，用时3ms

/**
 * 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==NULL)
            return result;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty())
        {
            int s=q.size();
            vector<int> inner;
            while(s--)
            {
                TreeNode* tmp=q.front();
                inner.push_back(tmp->val);
                if(tmp->left!=NULL)
                    q.push(tmp->left);
                if(tmp->right!=NULL)
                    q.push(tmp->right);
                q.pop();
            }
            result.push_back(inner);
        }
        return result;
    }
};

LeetCode107. Binary Tree Level Order Traversal II

题目链接：https://leetcode.com/problems/binary-tree-level-order-traversal-ii/?tab=Description

题目描述:

Given a binary tree, return the bottom-up level order traversal of its nodes' values. (ie, from left to right, level by level from leaf to root).

For example:
Given binary tree [3,9,20,null,null,15,7],

return its bottom-up level order traversal as:

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

思路：此题可以直接用上一题的方法，将结果逆序即可，也可将每一层在结果中插入的时候，一直在最前面插入

/**
 * 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>> levelOrderBottom(TreeNode* root) {
    vector<vector<int>> result;
        if(root==NULL)
            return result;
        queue<TreeNode*> q;
        q.push(root);
        while(!q.empty())
        {
            int s=q.size();
            vector<int> inner;
            while(s--)
            {
                TreeNode* tmp=q.front();
                inner.push_back(tmp->val);
                if(tmp->left!=NULL)
                    q.push(tmp->left);
                if(tmp->right!=NULL)
                    q.push(tmp->right);
                q.pop();
            }
            //result.insert(result.begin(),inner);
            result.push_back(inner);
        }
        reverse(result.begin(), result.end());
        return result;
    }
};