题意:
Given a binary tree, return the preorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3}
,
1 2 / 3
return [1,2,3]
.
Note: Recursive solution is trivial, could you do it iteratively?
思路:
写个非递归的前序遍历,用 stack.
代码:
C++:
/** * 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<int> preorderTraversal(TreeNode *root) { vector<int> ans; stack<TreeNode *> s; TreeNode *p = root; while(p != NULL || !s.empty()) { while(p != NULL) { ans.push_back(p->val); s.push(p); p = p->left; } if(!s.empty()) { p = s.top(); s.pop(); p = p->right; } } return ans; } };
Python:(python 写栈也是可以的)
# Definition for a binary tree node # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class Solution: # @param root, a tree node # @return a list of integers def preorderTraversal(self, root): ans = [] s = [] while root != None or s: while root != None: ans.append(root.val) s.append(root) root = root.left if s: root = s[-1] s.pop() root = root.right return ans
时间: 2024-11-03 01:35:55