Leetcode: Binary Tree Postorder Transversal

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].

Note: Recursive solution is trivial, could you do it iteratively?

难度：70

recursive方法很直接：

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

Iterative 最优做法：

pre-order traversal is root-left-right.

post-order traversal is left-right-root.

We can modify pre-order traversal to be root-right-left, and traverse the tree.

Finally, we reverse the output by the modified pre-order traversal to get post-order traversal.

 /**
  * Definition for a binary tree node.
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public List<Integer> postorderTraversal(TreeNode root) {
         LinkedList<Integer> res = new LinkedList<>();
         Stack<TreeNode> stack = new Stack<>();
         TreeNode p = root;
         while (p!=null || !stack.isEmpty()) {
             if (p != null) {
                 stack.push(p);
                 res.addFirst(p.val);
                 p = p.right;
             }
             else {
                 TreeNode node = stack.pop();
                 p = node.left;
             }
         }
         return res;
     }
 }

第一次Iterative的做法就没有那么strait forward的了，需要额外用一个ArrayList<TreeNode>来记录节点的访问情况

 /**
  * Definition for binary tree
  * public class TreeNode {
  *     int val;
  *     TreeNode left;
  *     TreeNode right;
  *     TreeNode(int x) { val = x; }
  * }
  */
 public class Solution {
     public List<Integer> postorderTraversal(TreeNode root) {
         ArrayList<Integer> res = new ArrayList<Integer>();
         if (root == null) return res;
         ArrayList<TreeNode> visited = new ArrayList<TreeNode>();
         LinkedList<TreeNode> stack = new LinkedList<TreeNode>();
         stack.push(root);
         while (root != null || !stack.isEmpty()) {
             if (root.left != null && !visited.contains(root.left)) {
                 stack.push(root.left);
                 root = root.left;
             }
             else if (root.right != null && !visited.contains(root.right)) {
                 stack.push(root.right);
                 root = root.right;
             }
             else {
                 visited.add(root);
                 res.add(stack.pop().val);
                 root = stack.peek();
             }
         }
         return res;
     }
 }