Binary Tree Level Order Traversal 题解
原创文章，拒绝转载
题目来源：https://leetcode.com/problems/binary-tree-level-order-traversal/description/
Description
Given a binary tree, return the level order traversal of its nodes‘ values. (ie, from left to right, level by level).
Example
For example:
Given binary tree [3,9,20,null,null,15,7],
3
/ 9 20
/ 15 7
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
Solution
class Solution {
public:
vector<vector<int> > levelOrder(TreeNode* root) {
vector<vector<int> > res;
if (root == NULL) return res;
queue<TreeNode*> q1, q2;
q2.push(root);
TreeNode* node;
int i;
while (!q2.empty()) {
while (!q2.empty()) {
node = q2.front();
q2.pop();
q1.push(node);
}
vector<int> level(q1.size());
i = 0;
while (!q1.empty()) {
node = q1.front();
q1.pop();
level[i++] = node -> val;
if (node -> left != NULL)
q2.push(node -> left);
if (node -> right != NULL)
q2.push(node -> right);
}
res.push_back(level);
}
return res;
}
};解题描述
这道题是经典的二叉树层次遍历问题，上面给出的解法使用2个队列，一开始将根节点放入队列q2，进入循环，每次将q2中全部节点出队加入队列q1，再将这些节点的子节点全部加入q2，即可实现层次遍历。
下面再给出一种讨论区的只使用单个队列实现的方法：
class Solution {
public:
vector<vector<int> > levelOrder(TreeNode* root) {
vector<vector<int> > res;
if (root == NULL) return res;
queue<TreeNode*> q;
q.push(root);
q.push(NULL); // 每一层结束的标志
vector<int> level;
while (!q.empty()) {
TreeNode* node = q.front();
q.pop();
if (node != NULL) {
level.push_back(node -> val);
if (node -> left != NULL)
q.push(node -> left);
if (node -> right != NULL)
q.push(node -> right);
} else { // 得到的节点是一层的结束标志NULL，则将当前层数组加入res
res.push_back(level);
level.clear();
if (!q.empty()) // 当前层已经扫描完毕，此时下一层也已经全部加入队列，需要往队列中加入层结束标志NULL
q.push(NULL);
}
}
return res;
}
};