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  • 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
    My 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]
    ]
    

     

    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;
        }
    };


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