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：类似于上一题。就是设置更新标志位，每一层反向遍历，递归的时候一直头插，迭代的时每层反向。

方法一：递归。

/**
* 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：非递归。

/**
* 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：堆栈。

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;
}
};