145. Binary Tree Postorder Traversal（二叉树后序遍历）

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

For example:
Given binary tree [1,null,2,3],

   1
    \
     2
    /
   3

return [3,2,1].

递归：

 class Solution {
     List<Integer> res = new ArrayList<Integer>();
     public List<Integer> postorderTraversal(TreeNode root) {
         help(root);
         return res;
     }
     private void help(TreeNode root){
         if(root == null) return ;
         help(root.left);
         help(root.right);
         res.add(root.val);
     }
 }

非递归：

终极版

 class Solution {
     public List<Integer> postorderTraversal(TreeNode root) {
         Stack<TreeNode> s =  new Stack<TreeNode>();
         List<Integer> res = new ArrayList<Integer>();
         while(root!=null||!s.isEmpty()){
             while(root!=null){
                 s.push(root);
                 res.add(0,root.val);
                 root = root.right;
             }

             if(!s.isEmpty()){
                 root = s.pop();
                 root = root.left;
             }
         }
         return res;
     }
 }

跟前序遍历差不多 ，stack 保存左子树，向右遍历，反向保存res.

 class Solution {
     List<Integer> res = new ArrayList<Integer>();
     public List<Integer> postorderTraversal(TreeNode root) {
         Stack<TreeNode> stack = new Stack<TreeNode>();
         TreeNode cur = root;
         while(cur!=null || !stack.isEmpty()){
             while(cur!=null){
                 res.add(0,cur.val);
                 stack.push(cur.left);
                 cur = cur.right;
             }
             cur = stack.pop();
         }
         return res;
     }
 }