题目：
二叉树的前序遍历
给出一棵二叉树，返回其节点值的前序遍历。
样例
给出一棵二叉树 {1,#,2,3},
1
\
2
/
3
返回 [1,2,3].
挑战
你能使用非递归实现么？
解题：
通过递归实现，根节点->左节点->右节点
Java程序：
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
* @param root: The root of binary tree.
* @return: Preorder in ArrayList which contains node values.
*/
public ArrayList<Integer> preorderTraversal(TreeNode root) {
// write your code here
ArrayList<Integer> res = new ArrayList<Integer>();
res = preTurn(res,root);
return res;
}
public ArrayList<Integer> preTurn(ArrayList<Integer> res ,TreeNode root){
if(root==null)
return res;
if(root!=null){
res.add(root.val);
if(root.left!=null){
res= preTurn(res,root.left);
}
if(root.right!=null){
res = preTurn(res,root.right);
}
}
return res;
}
}
总耗时: 1094 ms
Python程序：
"""
Definition of TreeNode:
class TreeNode:
def __init__(self, val):
self.val = val
self.left, self.right = None, None
""" class Solution:
"""
@param root: The root of binary tree.
@return: Preorder in ArrayList which contains node values.
"""
def preorderTraversal(self, root):
# write your code here
res = []
res = self.preorderTurn(res,root)
return res def preorderTurn(self,res,root):
if root==None:
return res
if root!=None:
res.append(root.val)
if root.left!=None:
res = self.preorderTurn(res,root.left)
if root.right!=None:
res = self.preorderTurn(res,root.right)
return res;
总耗时: 335 ms
非递归程序，直接来源
Java程序：
public class Solution {
/**
* @param root: The root of binary tree.
* @return: Preorder in ArrayList which contains node values.
*/
public ArrayList<Integer> preorderTraversal(TreeNode root) {
// write your code here
ArrayList<TreeNode> p = new ArrayList<TreeNode>();
ArrayList<Integer> res = new ArrayList<Integer>();
while (root != null || p.size() != 0){
res.add(root.val);
if (root.right != null)
p.add(root.right);
root = root.left;
if (root == null && p.size() != 0){
root = p.get(p.size()-1);
p.remove(p.size()-1);
}
}
return res;
}
}
总耗时: 1473 ms
Python程序：
class Solution:
"""
@param root: The root of binary tree.
@return: Preorder in ArrayList which contains node values.
"""
def preorderTraversal(self, root):
# write your code here
res = []
p = [root]
while root is not None or len(p) != 1:
res.append(root.val)
if root.right is not None:
p.append(root.right)
root = root.left
if root == None and len(p) != 1:
root = p[len(p) - 1]
del p[len(p) - 1]
return res
总耗时: 230 ms