Lintcode 二叉树的后序遍历

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]

二叉树后序遍历：左子树->右子树->根节点。下图二叉树的后序遍历结果为：edgfbca。

通过递归方式实现二叉树后序遍历的代码如下：
/**
* 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.
*/
ArrayList<Integer> list = new ArrayList<Integer>();
public ArrayList<Integer> postorderTraversal(TreeNode root) {
// write your code here
postorder(root);
return list;
}
private void postorder(TreeNode root){
if(root == null){
return;
}
postorder(root.left);
postorder(root.right);
list.add(root.val);
}
}
通过非递归方式实现二叉树的后序遍历的代码如下：
/**
* 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.
*/
ArrayList<Integer> list = new ArrayList<Integer>();
Stack<TreeNode> stack = new Stack<TreeNode>();
public ArrayList<Integer> postorderTraversal(TreeNode root) {
// write your code here
postorder(root);
return list;
}
private void postorder(TreeNode root){
TreeNode p = root;
while(root!=null){
while(root.left!=null){
stack.push(root);
root = root.left;
}
while(root!=null&&(root.right==null||root.right == p)){
list.add(root.val);
p = root;
if(stack.isEmpty()) return;
root = stack.pop();
}
stack.push(root);
root = root.right;
}
}
}