C++实现LeetCode(102.二叉树层序遍历)
[LeetCode] 102. 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，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，参见代码如下：
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;
}
};
下面来看递归的写法，核心就在于需要一个二维数组，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);
}
};
Github 同步地址：
https://github.com/grandyang/leetcode/issues/102
类似题目：
Binary Tree Level Order Traversal II
Binary Tree Zigzag Level Order Traversal
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/
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