Leetcode 树 Binary Tree Zigzag Level Order Traversal

本文为senlie原创，转载请保留此地址：http://blog.csdn.net/zhengsenlie

Binary Tree Zigzag Level Order Traversal

Total Accepted: 8678 Total
Submissions: 33047My Submissions

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.

题意：给定一棵二叉树，返回按zigzag层次遍历的结果

思路：

还是跟前面的Binary Tree Level Order Traversal的思路一样

即从上往下按层遍历二叉树，将每一层的节点存放到该层对应的数组中

最后将得到的总数组中奇数层（从0层开始计数）的子数组reverse一下就可以了

复杂度：时间O(n)，空间O(n)

相关题目：

Binary Tree Level Order Traversal

Binary Tree Level Order Traversal II

/**
 * 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> > result;
    	queue<NodeWithLevel> q;
    	q.push(NodeWithLevel(root, 0));
    	while (!q.empty())
    	{
    		NodeWithLevel cur = q.front(); q.pop();
    		TreeNode *p = cur.p;
    		if(p){
    			if(result.size() <= cur.level){
    				vector<int> tem;
    				tem.push_back(p->val);
    				result.push_back(tem);

    			}else{
    				result[cur.level].push_back(p->val);
    			}
    			NodeWithLevel left(p->left, cur.level + 1);
    			NodeWithLevel right(p->right, cur.level + 1);
    			q.push(left);
    			q.push(right);
    		}
    	}
    	for(int i = 1; i < result.size(); i += 2) reverse(result[i].begin(), result[i].end());
    	return result;
    }

private:
    struct NodeWithLevel{
    	TreeNode *p;
    	int level;
    	NodeWithLevel(TreeNode *pp, int l):p(pp), level(l){}
    };

};

Leetcode 树 Binary Tree Zigzag Level Order Traversal