題目鏈接
tag:
- Medium丁存；
- queue肩杈；
question：
??Given a binary tree, return the zigzag level order traversal of its nodes' values. (ie, from left to right, then right to left for the next level and alternate between).
Example:
Given binary tree [3,9,20,null,null,15,7],
3
/
9 20
/
15 7
return its zigzag level order traversal as:
[
[3],
[20,9],
[15,7]
]
思路：
??本題和102. 二叉樹(shù)層序遍歷類(lèi)似。通過(guò)一個(gè)隊(duì)列來(lái)控制解寝，仍然是當(dāng)做一個(gè)普通的 BFS 來(lái)實(shí)現(xiàn)扩然。唯一不同的是，判斷是否需要將當(dāng)前的數(shù)組反轉(zhuǎn)聋伦。在程序中定義一個(gè)標(biāo)簽（flag）來(lái)實(shí)現(xiàn)夫偶。flag代表樹(shù)的深度，根節(jié)點(diǎn)為零觉增。樹(shù)的深度為奇數(shù)的時(shí)候兵拢，不需要反轉(zhuǎn)，樹(shù)的深度為偶數(shù)的時(shí)候逾礁，需要反轉(zhuǎn)说铃。代碼如下：
/**
* 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>> zigzagLevelOrder(TreeNode* root) {
if (!root) return {};
vector<vector<int>> res;
queue<TreeNode*> q;
q.push(root);
int flag = 0; // tree depth
while (!q.empty()) {
vector<int> level;
for (int i=q.size(); i>0; --i) {
TreeNode *tmp = q.front();
q.pop();
level.push_back(tmp->val);
if (tmp->left) q.push(tmp->left);
if (tmp->right) q.push(tmp->right);
}
++flag;
if (flag & 1)
res.push_back(level);
else {
reverse(level.begin(), level.end());
res.push_back(level);
}
}
return res;
}
};
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