【leetcode】Binary Tree Zigzag Level Order Traversal

最新推荐文章于 2025-11-25 14:23:57 发布

最新推荐文章于 2025-11-25 14:23:57 发布 · 74 阅读

文章标签：

#c/c++

本文介绍了一种解决二叉树锯齿形层序遍历问题的方法，利用两个栈交替存储节点，实现从左到右再从右到左的锯齿形遍历。通过一个具体的示例展示了如何获取二叉树各层节点值并按要求返回。

Question ：

Given a binary tree, return thezigzag level ordertraversal 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]
]

confused what"{1,#,2,3}"means?> read more on how binary tree is serialized on OJ.

Anwser 1 ：

/**
 * 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>> ret;
        if(root == NULL) return ret;
        
        vector<int> vec;
        stack<TreeNode *> Q;
        stack<TreeNode *> Q2;   // extra space
        bool flag = true;
        
        Q.push(root);
        while(!Q.empty()){
            TreeNode *tmp = Q.top();
            Q.pop();
            
            if(tmp != NULL){
                vec.push_back(tmp->val);
                if(flag){
                    if(tmp->left) Q2.push(tmp->left);
                    if(tmp->right) Q2.push(tmp->right);
                } else {
                    if(tmp->right) Q2.push(tmp->right);
                    if(tmp->left) Q2.push(tmp->left);
                }
            }
            
            if(Q.empty()){      // one row end
                ret.push_back(vec);
                vec.clear();
                flag = !flag;
                swap(Q, Q2);
            }
        }
        return ret;
    }
};