树——binary-tree-postorder-traversal（树的后序遍历）

问题：

　　Given a binary tree, return the postorder traversal of its nodes' values.

　　For example:
　　Given binary tree{1,#,2,3},

   　　1
    　　\
   　　  2
  　　  /
  　　 3

　　return[3,2,1].

递归方法：

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 import java.util.*;
 public class Solution {
     public ArrayList<Integer> postorderTraversal(TreeNode root) {
         ArrayList<Integer> list = new ArrayList<Integer>();
         if(root==null)
             return list;
         postorderTraversal(root, list);
         return list;
     }
     public void postorderTraversal(TreeNode root, ArrayList<Integer> list){
         if(root==null)
             return;

         if(root.left!=null)
             postorderTraversal(root.left, list);   //先遍历左节点
         if(root.right!=null)
             postorderTraversal(root.right, list);   //再遍历右节点

  　　　　list.add(root.val);    //最后遍历根节点
     } 
 }

非递归方法（利用栈）

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 import java.util.*;
 public class Solution {
     public ArrayList<Integer> postorderTraversal(TreeNode root) {
        ArrayList<Integer> list = new ArrayList<Integer>();
         if(root==null)
             return list;
         Stack<TreeNode> stack = new Stack<TreeNode>();
         Stack<TreeNode> st = new Stack<TreeNode>();
         stack.push(root);
         while(!stack.isEmpty()){
             TreeNode node = stack.pop();
             if(node.left!=null)
                 stack.push(node.left);
             if(node.right!=null)
                 stack.push(node.right);
             st.push(node);
         }
         while(!st.isEmpty()){
             list.add(st.pop().val);
         }
         return list;
     }

 }