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).
For example:
Given binary tree {3,9,20,#,#,15,7},
3
/ \
9 20
/ \
15 7
return its zigzag level order traversal as:
[ [3], [20,9], [15,7] ]
confused what "{1,#,2,3}" means? > read more on how binary tree is serialized on OJ.
这道二叉树的之字形层序遍历是之前那道[LeetCode] Binary Tree Level Order Traversal 二叉树层序遍历的变形，不同之处在于一行是从左到右遍历，下一行是从右往左遍历，交叉往返的之字形的层序遍历。根据其特点我们用到栈的后进先出的特点，这道题我们维护两个栈，相邻两行分别存到两个栈中，进栈的顺序也不相同，一个栈是先进左子结点然后右子节点，另一个栈是先进右子节点然后左子结点，这样出栈的顺序就是我们想要的之字形了，代码如下：
/** * Definition for binary tree * 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) { vector<vector<int> >res; if (!root) return res; stack<TreeNode*> s1; stack<TreeNode*> s2; s1.push(root); vector<int> out; while (!s1.empty() || !s2.empty()) { while (!s1.empty()) { TreeNode *cur = s1.top(); s1.pop(); out.push_back(cur->val); if (cur->left) s2.push(cur->left); if (cur->right) s2.push(cur->right); } if (!out.empty()) res.push_back(out); out.clear(); while (!s2.empty()) { TreeNode *cur = s2.top(); s2.pop(); out.push_back(cur->val); if (cur->right) s1.push(cur->right); if (cur->left) s1.push(cur->left); } if (!out.empty()) res.push_back(out); out.clear(); } return res; } };
比如对于题干中的那个例子：
3
/ \
9 20
/ \
15 7
我们来看每一层两个栈s1, s2的情况：
s1: 3
s2:
s1:
s2: 9 20
s1: 7 15
s2:
本文转自博客园Grandyang的博客，原文链接：二叉树的之字形层序遍历[LeetCode] Binary Tree Zigzag Level Order Traversal ，如需转载请自行联系原博主。
