[LeetCode] Binary Tree Level Order Traversal 二叉树层序遍历

 

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]
]

层序遍历二叉树是典型的广度优先搜索BFS的应用,但是这里稍微复杂一点的是,我们要把各个层的数分开,存到一个二维向量里面,大体思路还是基本相同的,建立一个queue,然后先把根节点放进去,这时候找根节点的左右两个子节点,这时候去掉根节点,此时queue里的元素就是下一层的所有节点,用一个for循环遍历它们,然后存到一个一维向量里,遍历完之后再把这个一维向量存到二维向量里,以此类推,可以完成层序遍历。代码如下:

 

解法一:

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        if (!root) return {};
        vector<vector<int>> res;
        queue<TreeNode*> q{{root}};
        while (!q.empty()) {
            vector<int> oneLevel;
            for (int i = q.size(); i > 0; --i) {
                TreeNode *t = q.front(); q.pop();
                oneLevel.push_back(t->val);
                if (t->left) q.push(t->left);
                if (t->right) q.push(t->right);
            }
            res.push_back(oneLevel);
        }
        return res;
    }
};

 

下面我们来看递归的写法,核心就在于我们需要一个二维数组,和一个变量level,关于level的作用可以参见博主的另一篇博客 Binary Tree Level Order Traversal II 中的讲解,参见代码如下:

 

解法二:

class Solution {
public:
    vector<vector<int>> levelOrder(TreeNode* root) {
        vector<vector<int>> res;
        levelorder(root, 0, res);
        return res;
    }
    void levelorder(TreeNode* node, int level, vector<vector<int>>& res) {
        if (!node) return;
        if (res.size() == level) res.push_back({});
        res[level].push_back(node->val);
        if (node->left) levelorder(node->left, level + 1, res);
        if (node->right) levelorder(node->right, level + 1, res);
    }
};

 

类似题目:

Binary Tree Level Order Traversal II

Binary Tree Zigzag Level Order Traversal

Minimum Depth of Binary Tree

Binary Tree Vertical Order Traversal 

Average of Levels in Binary Tree

N-ary Tree Level Order Traversal

 

参考资料:

https://leetcode.com/problems/binary-tree-level-order-traversal/

https://leetcode.com/problems/binary-tree-level-order-traversal/discuss/33445/Java-Solution-using-DFS

https://leetcode.com/problems/binary-tree-level-order-traversal/discuss/33450/Java-solution-with-a-queue-used

https://leetcode.com/problems/binary-tree-level-order-traversal/discuss/114449/A-general-approach-to-level-order-traversal-questions-in-Java

 

    原文作者:Grandyang
    原文地址: http://www.cnblogs.com/grandyang/p/4051321.html
    本文转自网络文章,转载此文章仅为分享知识,如有侵权,请联系博主进行删除。
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