[LeetCode] Binary Tree Zigzag Level Order Traversal 二叉树的之字形层序遍历

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:

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