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;
    }
}