标签：leetcode   算法   

问题描述：

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.

代码：

vector<vector<int> > zigzagLevelOrder(TreeNode *root) {  //C++
        vector<vector<int> > result;
        if(root == NULL)
            return result;
            
        int toleft = 1;
        stack<TreeNode*> left;
        stack<TreeNode*> right;
        
        left.push(root);
        while(!(left.empty()&&right.empty()))
        {
            vector<int> vec;
            if(toleft==1){
                while(!left.empty()){
                    TreeNode* temp = left.top();
                    vec.push_back(temp->val);
                    if(temp->left != NULL)
                        right.push(temp->left);
                    if(temp->right != NULL)
                        right.push(temp->right);
                    left.pop();
                }
                result.push_back(vec);
            }
            else{
                while(!right.empty()){
                    TreeNode* temp = right.top();
                    vec.push_back(temp->val);
                    if(temp->right != NULL)
                        left.push(temp->right);
                    if(temp->left != NULL)
                        left.push(temp->left);
                    right.pop();
                }
                result.push_back(vec);
            }
        toleft *=-1;
        }
    return result;
    }

[leetcode]103 Binary Tree Zigzag Level Order Traversal

标签：leetcode   算法   

原文地址：http://blog.csdn.net/chenlei0630/article/details/42681253