IT工程師數位筆記本

之字形層次遍歷二叉樹：層次遍歷二叉樹，并且奇數行為從左到右（根節點為第一層），偶數行為從右到左。 先寫一個容易實現的，參照《編程之美》3.10分層遍歷二叉樹的做法，先編寫一個用來處理制定層的函數，然后逐層調用這個函數即可。

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.

OJ's Binary Tree Serialization: The serialization of a binary tree follows a level order traversal, where '#' signifies a path terminator where no node exists below. Here's an example: 1 /
2 3 / 4
5 The above binary tree is serialized as"{1,2,3,#,#,4,#,#,5}".

/**
 * 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;
		for (int level = 0;; level++){
			vector<int> q = ZigzagNodeAtLevel(root, level);
			if (q.size() == 0){
				break;
			}
			result.push_back(q);
		}
        return result;
	}

	vector<int> ZigzagNodeAtLevel(TreeNode* root, int level){
		vector<int> res;
		if (!root || level<0){
			return res;
		}
		if (level == 0){
			res.push_back(root->val);
		}
		// return PrintNodeAtLevel(root->left, level - 1) + PrintNodeAtLevel(root->right, level - 1);
		vector<int> left_res = ZigzagNodeAtLevel(root->left, level - 1);
		vector<int> right_res = ZigzagNodeAtLevel(root->right, level - 1);
        
        if(level%2==1){
            for (int i = right_res.size()-1; i>=0; i--){
                res.push_back(right_res[i]);
            }
            for (int i = left_res.size()-1; i >=0; i--){
                res.push_back(left_res[i]);
            }

        }else{
            for (int i = left_res.size()-1; i >=0; i--){
                res.push_back(left_res[i]);
            }
            for (int i = right_res.size()-1; i>=0; i--){
                res.push_back(right_res[i]);
            }
        }
		
		return res;
	}
};