给定一个二叉树,返回其节点值的锯齿形层次遍历。(即先从左往右,再从右往左进行下一层遍历,以此类推,层与层之间交替进行)。
例如:
给定二叉树 [3,9,20,null,null,15,7],
3
/ \
9 20
/ \
15 7
返回锯齿形层次遍历如下:
[
[3],
[20,9],
[15,7]
]
这个题目与普通的层次遍历思想没有多大区别,这是这里需要两个方向,所有设置两个栈或者队列,这里使用了栈,每个栈保存一层的节点,然后从一个方向输出,另一个栈保存另一层节点,保存时就反向保存,然后从另一个方向输出。
C++源代码:
/**
* 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>> res;
if (!root) return res;
vector<int> out;
stack<TreeNode*> s1;
stack<TreeNode*> s2;
s1.push(root);
while(!s1.empty() || !s2.empty()){
while(!s1.empty()){
TreeNode *tmp = s1.top();
s1.pop();
out.push_back(tmp->val);
if(tmp->left) s2.push(tmp->left);
if(tmp->right) s2.push(tmp->right);
}
if(!out.empty()) res.push_back(out);
out.clear();
while(!s2.empty()){
TreeNode *tmp = s2.top();
s2.pop();
out.push_back(tmp->val);
if(tmp->right) s1.push(tmp->right);
if(tmp->left) s1.push(tmp->left);
}
if(!out.empty()) res.push_back(out);
out.clear();
}
return res;
}
};
python3源代码:
# Definition for a binary tree node.
# class TreeNode:
# def __init__(self, x):
# self.val = x
# self.left = None
# self.right = None
class Solution:
def zigzagLevelOrder(self, root: 'TreeNode') -> 'List[List[int]]':
res = []
if root==None: return res
out = []
s1 = [root]
s2 = []
while len(s1)!=0 or len(s2)!=0:
while len(s1)!=0:
tmp = s1.pop()
out.append(tmp.val)
if tmp.left: s2.append(tmp.left)
if tmp.right: s2.append(tmp.right)
if len(out)!=0: res.append(out)
out = []
while len(s2)!=0:
tmp = s2.pop()
out.append(tmp.val)
if tmp.right: s1.append(tmp.right)
if tmp.left: s1.append(tmp.left)
if len(out)!=0: res.append(out)
out = []
return res