LeetCode算法题103：二叉树的锯齿形层次遍历解析

给定一个二叉树，返回其节点值的锯齿形层次遍历。（即先从左往右，再从右往左进行下一层遍历，以此类推，层与层之间交替进行）。

例如：
给定二叉树 [3,9,20,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7
返回锯齿形层次遍历如下：

[
  [3],
  [20,9],
  [15,7]
]

这个题目与普通的层次遍历思想没有多大区别，这是这里需要两个方向，所有设置两个栈或者队列，这里使用了栈，每个栈保存一层的节点，然后从一个方向输出，另一个栈保存另一层节点，保存时就反向保存，然后从另一个方向输出。

C++源代码：

/**
 * 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>> res;
        if (!root) return res;
        vector<int> out;
        stack<TreeNode*> s1;
        stack<TreeNode*> s2;
        s1.push(root);
        while(!s1.empty() || !s2.empty()){
            while(!s1.empty()){
                TreeNode *tmp = s1.top();
                s1.pop();
                out.push_back(tmp->val);
                if(tmp->left) s2.push(tmp->left);
                if(tmp->right) s2.push(tmp->right);
            }
            if(!out.empty()) res.push_back(out);
            out.clear();
            while(!s2.empty()){
                TreeNode *tmp = s2.top();
                s2.pop();
                out.push_back(tmp->val);
                if(tmp->right) s1.push(tmp->right);
                if(tmp->left) s1.push(tmp->left);
            }
            if(!out.empty()) res.push_back(out);
            out.clear();
        }
        return res;
    }
};

python3源代码：

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def zigzagLevelOrder(self, root: 'TreeNode') -> 'List[List[int]]':
        res = []
        if root==None: return res
        out = []
        s1 = [root]
        s2 = []
        while len(s1)!=0 or len(s2)!=0:
            while len(s1)!=0:
                tmp = s1.pop()
                out.append(tmp.val)
                if tmp.left: s2.append(tmp.left)
                if tmp.right: s2.append(tmp.right)
            if len(out)!=0: res.append(out)
            out = []
            while len(s2)!=0:
                tmp = s2.pop()
                out.append(tmp.val)
                if tmp.right: s1.append(tmp.right)
                if tmp.left: s1.append(tmp.left)
            if len(out)!=0: res.append(out)
            out = []
        return res