题目：
二叉树的中序遍历
给出一棵二叉树,返回其中序遍历
样例
给出二叉树 {1,#,2,3},
1
\
2
/
3
返回 [1,3,2].
挑战
你能使用非递归算法来实现么?
解题：
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: Inorder in ArrayList which contains node values.
*/
public ArrayList<Integer> inorderTraversal(TreeNode root) {
// write your code here
ArrayList<TreeNode> p = new ArrayList<TreeNode>();
ArrayList<Integer> res = new ArrayList<Integer>();
while(root != null || p.size() != 0){
while(root != null){
p.add(root);
root = root.left;
}
root = p.get(p.size()-1);
p.remove(p.size()-1);
res.add(root.val);
root = root.right;
}
return res;
} }
总耗时: 1238 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: Inorder in ArrayList which contains node values.
"""
def inorderTraversal(self, root):
# write your code here
p = [root]
res = [0]
while root is not None or len(p) != 1:
while root is not None:
p.append(root)
root = root.left
root = p[len(p)-1]
del p[len(p)-1]
res.append(root.val)
root = root.right
n = len(res)
return res[1:n]
总耗时: 263 ms
非递归程序，理解不透，还需要人丑就要多读书
根据上面灵感，递归程序如下：
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: Inorder in ArrayList which contains node values.
*/
public ArrayList<Integer> inorderTraversal(TreeNode root) {
// write your code here ArrayList<Integer> res = new ArrayList<Integer>();
res = inorderTrun(res,root);
return res;
}
public ArrayList<Integer> inorderTrun(ArrayList<Integer> res,TreeNode root){
if(root == null)
return res;
if(root!=null){
if(root.left!=null){
res = inorderTrun(res,root.left);
}
res.add(root.val);
if(root.right!=null){
res = inorderTrun(res,root.right);
}
}
return res;
} }
总耗时: 1714 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: Inorder in ArrayList which contains node values.
"""
def inorderTraversal(self, root):
# write your code here
res = []
res = self.inorderTrun(res,root)
return res def inorderTrun(self,res,root):
if root==None:
return res
if root.left!=None:
res = self.inorderTrun(res,root.left)
res.append(root.val)
if root.right!=None:
res = self.inorderTrun(res,root.right)
return res
总耗时: 213 ms
根据上面的程序理解，可根据栈实现，上面定义的ArrayList也是起到栈的作用