LeetCode:Binary Tree Zigzag Level Order Traversal

题目链接

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]
]                                                                          本文地址

分析：首先肯定的是要对树进行层序遍历，但是相邻两层的元素遍历顺序是相反的，因此传统的非递归遍历方法用一个队列肯定是无法实现。我们用两个栈来实现，同一层元素都在同一个栈中，相邻的两层元素放在不同的栈中，比如第一层元素放在第一个栈，那么第二层就放在第二个栈，第三层就放在第三个栈......如果某一层元素是从左往右遍历的，那么这层元素的孩子节点入栈顺序就是先左孩子后右孩子，相反如果某一层元素是从右往左遍历的，入栈顺序就是先右孩子后左孩子。

/**
* 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) {
// IMPORTANT: Please reset any member data you declared, as
// the same Solution instance will be reused for each test case.
vector<vector<int> >res;
if(root == NULL)return res;
stack<TreeNode*> S,S2;
S.push(root);
bool isS = true;
vector<int>tmpres;
while(S.empty() == false || S2.empty() == false)
{
TreeNode *p;
if(isS)
{

p = S.top();
S.pop();
tmpres.push_back(p->val);
if(p->left)S2.push(p->left);
if(p->right)S2.push(p->right);
if(S.empty() == true)
{
res.push_back(tmpres);
tmpres.clear();
isS = false;
}
}
else
{
p = S2.top();
S2.pop();
tmpres.push_back(p->val);
if(p->right)S.push(p->right);
if(p->left)S.push(p->left);
if(S2.empty() == true)
{
res.push_back(tmpres);
tmpres.clear();
isS = true;
}
}
}
return res;
}
};

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