[LeetCode] 103. 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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

 

return its zigzag level order traversal as:

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

102. Binary Tree Level Order Traversal 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。

Java：

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/** * Definition for binary tree * public class TreeNode { * int val; * TreeNode left; * TreeNode right; * TreeNode(int x) { val = x; } * } */ public class Solution {     public List<List<Integer>> zigzagLevelOrder(TreeNode root) {         List<List <Integer>> result = new ArrayList<List<Integer>>();         ArrayList<TreeNode> q = new ArrayList<TreeNode>();         ArrayList<Integer> level = new ArrayList<Integer>();         if (root == null){             return result;         }          q.add(root);         level.add(0);         while (q.size() > 0){             TreeNode node = q.remove(0);             int cl = level.remove(0);             if (result.size() <= cl){                 result.add(new ArrayList<Integer>());             }             if (cl % 2 == 0){                 result.get(cl).add(node.val);             }             else{                 result.get(cl).add(0, node.val);             }             if (node.left != null){                 q.add(node.left);                 level.add(cl + 1);             }             if (node.right != null){                 q.add(node.right);                 level.add(cl + 1);             }         }//while         return result;     }//zigzag }

Python：

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class TreeNode:     def __init__(self, x):         self.val = x         self.left = None         self.right = None   class Solution:     # @param root, a tree node     # @return a list of lists of integers     def zigzagLevelOrder(self, root):         if root is None:             return []         result, current, level = [], [root], 1         while current:             next_level, vals = [], []             for node in current:                 vals.append(node.val)                 if node.left:                     next_level.append(node.left)                 if node.right:                     next_level.append(node.right)             if level % 2:                 result.append(vals)             else:                 result.append(vals[::-1])             level += 1             current = next_level         return result

C++：

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/**  * 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;     } };

　　　　

类似题目：

[LeetCode] 102. Binary Tree Level Order Traversal 二叉树层序遍历

 

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