Binary Tree Zigzag Level Order Traversal (LeetCode) 层序遍历二叉树

题目描述:

Binary Tree Zigzag Level Order Traversal

 
AC Rate: 399/1474
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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},

    3
   / 
  9  20
    /  
   15   7

 

return its zigzag level order traversal as:

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

 

代码:

/**
 * 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
        queue<TreeNode *> qu;
        qu.push(root);
        qu.push(NULL);  // flag for end of one level
        
        bool right2left = false;
        vector<vector<int> > res;
        
        if(NULL==root) return res;
        
        vector<int> lev;  // for nodes in one level
    
        while(1)
        {
            TreeNode *cur = qu.front();
            qu.pop();
            
            if(cur)
            {
                lev.push_back(cur->val);
                if(cur->left) qu.push(cur->left);
                if(cur->right) qu.push(cur->right);
            }
            else // end of one level
            {
                if(!right2left)
                {
                    res.push_back(lev);
                }
                else
                {
                    reverse(lev.begin(),lev.end());
                    res.push_back(lev);
                }
                right2left = !right2left;
                
                if(qu.size()==0) break;
                
                qu.push(NULL);
                lev.clear();
            }
        }
        return res;
    }
};


原文地址:https://www.cnblogs.com/riskyer/p/3347909.html