题目：
二叉树的后序遍历
给出一棵二叉树，返回其节点值的后序遍历。
样例
给出一棵二叉树 {1,#,2,3},
1
\
2
/
3
返回 [3,2,1]
挑战
你能使用非递归实现么？
解题：
递归程序好简单
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: Postorder in ArrayList which contains node values.
*/
public ArrayList<Integer> postorderTraversal(TreeNode root) {
// write your code here
ArrayList<Integer> res = new ArrayList<Integer>();
res = postorder(res,root);
return res;
}
public ArrayList<Integer> postorder(ArrayList<Integer> res,TreeNode root){
if(root==null)
return res;
if(root.left!=null)
res = postorder(res,root.left);
if(root.right!=null)
res = postorder(res,root.right);
res.add(root.val);
return res;
}
}
总耗时: 1210 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: Postorder in ArrayList which contains node values.
"""
def postorderTraversal(self, root):
# write your code here
res = []
res = self.postorder(res,root)
return res def postorder(self,res,root):
if root==None:
return res
if root.left!=None:
res = self.postorder(res,root.left)
if root.right!=None:
res = self.postorder(res,root.right)
res.append(root.val)
return res
总耗时: 380 ms
非递归程序，直接来源
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: Postorder in ArrayList which contains node values.
*/
public ArrayList<Integer> postorderTraversal(TreeNode root) {
// write your code here
int a = 1;
ArrayList<TreeNode> s = new ArrayList<TreeNode>();
ArrayList<Integer> res = new ArrayList<Integer>();
if (root == null) return res;
while(a == 1){
while(root.left != null || root.right != null){
if (root.left != null){
s.add(root);
root = root.left;
}
else{
s.add(root);
root = root.right;
}
}
TreeNode y = s.get(s.size()-1);
while (root == y.right || y.right == null){
res.add(root.val);
s.remove(s.size()-1);
if (s.size() == 0){
a = 0;
res.add(y.val);
break;
}
root = y;
y = s.get(s.size()-1);
}
if (root == y.left && y.right != null){
res.add(root.val);
root = y.right;
}
}
return res;
}
}
总耗时: 1388 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: Postorder in ArrayList which contains node values.
"""
def postorderTraversal(self, root):
# write your code here
a = 1
s = [root]
res = []
if root is None:
return res[1:1]
while a == 1:
while root.left is not None or root.right is not None:
if root.left is not None:
s.append(root)
root = root.left
else:
s.append(root)
root = root.right
y = s[len(s)-1]
while root == y.right or y.right is None:
res.append(root.val)
del s[len(s)-1]
if len(s) == 1:
a = 0
res.append(y.val)
break
root = y
y = s[len(s)-1]
if root == y.left and y.right is not None:
res.append(root.val)
root = y.right
return res
总耗时: 360 ms