leetcode_question_103 Binary Tree Zigzag Level Order Traversal

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最新推荐文章于 2025-08-19 21:22:03 发布

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分类专栏：程序员笔试面试 Leetcode 算法与数据结构文章标签： leetcode Binary Tree Zigzag L bfs

本文链接：https://blog.youkuaiyun.com/doc_sgl/article/details/11937835

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本文介绍了一种算法，用于实现二叉树节点值的锯齿形层序遍历。从左到右再从右到左交替进行，并提供了一个具体的C++实现示例。

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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},

return its zigzag level order traversal as:

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

BFS:

/**
 * 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) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        vector<vector<int>> matrix;
    	if(root == NULL) return matrix;
		vector<int> tmp;
		tmp.push_back(root->val);
		matrix.push_back(tmp);
		vector<TreeNode*> path;
		path.push_back(root);
		int count = 1;
		bool lefttoright = false;
		while(!path.empty()){
			if(path[0]->left) path.push_back(path[0]->left);
			if(path[0]->right) path.push_back(path[0]->right);
			path.erase(path.begin());
			count--;
			if(count == 0 && path.size())
			{
				vector<int> tmp;
				if(lefttoright){
					vector<TreeNode*>::iterator it = path.begin();
					for(; it != path.end(); ++it)
						tmp.push_back((*it)->val);
					lefttoright = false;
				}else{
					vector<TreeNode*>::iterator it = path.end();
					for(--it;it>=path.begin();--it)
						tmp.push_back((*it)->val);
					lefttoright = true;
				}
				matrix.push_back(tmp);
				count = path.size();
			}
		}
		return matrix;
    }
};