BinaryTreePostorderTrave.java

/*
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].
*/

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
// iterative solution 9/14/2014
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        ArrayList<Integer> result = new ArrayList<Integer>();
        if (root == null){
            return result;
        }
        
        Stack<TreeNode> stack =new Stack<TreeNode> ();
        LinkedList<TreeNode> visitedList= new LinkedList<TreeNode> ();
        
        while(root!=null){
            stack.push(root);
            root=root.left;
        }
        
        while(!stack.isEmpty()){
            TreeNode current = stack.peek();
            if (current.right == null || visitedList.contains(current)){
                result.add(current.val);
                stack.pop();
                if (visitedList.contains(current)){
                    visitedList.remove(current);
                }
            }
            else{
                visitedList.add(current);
                root=current.right;
                
                while (root!=null){
                    stack.push(root);
                    root=root.left;
                }
                
            }
        }
        
        return result;
    }
} 
 // Iterative Solution
public class Solution {
    public List<Integer> postorderTraversal(TreeNode root) {
        
        List<Integer> result=new ArrayList<Integer>();
        if (root==null){
            return result;
        }
        
        Stack<TreeNode> stack=new Stack<TreeNode>();
        TreeNode runner=root;
        
        while (runner!=null){
            stack.push(runner);
            runner=runner.left;
        }
        // child node used to record if mid node's child has been visited inorder to tell if need to pop out        
        //the mid node
        
        TreeNode child=null;
        
        while (!stack.isEmpty()){
            TreeNode current=stack.peek();
            if (current.right==null || current.right==child){
                result.add(stack.pop().val);
                // catch the child node, inorder mid node could check if its child already be visited
                child=current;
            }
            else{
                current=current.right;
                while (current!=null){
                    stack.push(current);
                    current=current.left;
                }
            }
        }
        
        return result; 
    }
}


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