LeetCode--Binary Tree Zigzag Level Order Traversal

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,null,null,15,7],

    3
   / \
  9  20
    /  \
   15   7

return its zigzag level order traversal as:

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

思路：類似於上一題。就是設置更新標誌位，每一層反向遍歷，遞歸的時候一直頭插，迭代的時每層反向。

方法一：遞歸。

/**
 * Definition for a binary tree node.
 * 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) {
        vector<vector<int>>result;
        traverse(root,1,result,true);
        return result;
    }
    void traverse(TreeNode* root,int level,vector<vector<int>>&result,bool left_to_right){
        if(!root) return;
        if(level>result.size()){
            result.push_back(vector<int>());
        }
        if(left_to_right){
            result[level-1].push_back(root->val);
        }
        else{
            result[level-1].insert(result[level-1].begin(),root->val);
        }
        traverse(root->left,level+1,result,!left_to_right);
        traverse(root->right,level+1,result,!left_to_right);
    }
};

方法二：非遞歸。

/**
 * Definition for a binary tree node.
 * 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) {
        vector<vector<int>>result;
        queue<TreeNode*>current,next;
        bool left_to_right=true;
        if(!root) return result;
        else current.push(root);
        while(!current.empty()){
            vector<int>level;
            while(!current.empty()){
                TreeNode *node=current.front();
                current.pop();
                level.push_back(node->val);
                if(node->left) next.push(node->left);
                if(node->right) next.push(node->right);
            }
            if(!left_to_right) reverse(level.begin(),level.end());
            result.push_back(level);
            left_to_right=!left_to_right;
            swap(current,next);
        }
        return result;
    }
};

方法三：堆棧。

class Solution {
public:
    vector<vector<int> > Print(TreeNode* pRoot) {
        vector<vector<int> > result;
        stack<TreeNode *> stack1,stack2;
        if(pRoot)
            stack1.push(pRoot);
        TreeNode *node;
        while(!stack1.empty() || !stack2.empty()){
            vector<int> data;
            if(!stack1.empty()){
                while(!stack1.empty()){
                    node = stack1.top();
                    stack1.pop();
                    data.push_back(node->val);
                    if(node->left!=NULL)
                        stack2.push(node->left);
                    if(node->right!=NULL)
                        stack2.push(node->right);
                }
                result.push_back(data);
            }
            else if(!stack2.empty()){
                while(!stack2.empty()){
                    node = stack2.top();
                    stack2.pop();
                    data.push_back(node->val);
                    if(node->right!=NULL)
                        stack1.push(node->right);
                    if(node->left!=NULL)
                        stack1.push(node->left);
                }
                result.push_back(data);
            }
        }
        return result;
    }
};