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;
    }
}