【Leetcode】【Medium】Binary Tree Zigzag Level Order Traversal

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]
]

解题思路：

按层遍历，建立两个栈，一个保存当前结点，一个按照Z型保存下一层结点。

使用一个bool变量来标注每次读取子节点的方向；

代码：

/**
* 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) {
vector<vector<int> > ret;
stack<TreeNode*> cur_layer;
stack<TreeNode*> next_layer;
cur_layer.push(root);
bool dirt = false;

if (!root)
return ret;

while (!cur_layer.empty()) {
vector<int> layer_val;
while (!cur_layer.empty()) {
TreeNode* node = cur_layer.top();
cur_layer.pop();
layer_val.push_back(node->val);
if (dirt) {
if (node->right)
next_layer.push(node->right);
if (node->left)
next_layer.push(node->left);
} else {
if (node->left)
next_layer.push(node->left);
if (node->right)
next_layer.push(node->right);
}
}
ret.push_back(layer_val);
swap(cur_layer, next_layer);
dirt = !dirt;
}
return ret;
}
};