[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：

/**
* 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：

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++：

/**
 * 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;
    }
};

　　　　

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

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