145. Binary Tree Postorder Traversal QuestionEditorial Solution

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?

每次都先push right， 然后left。 但是需要一个sentinel存每次上一次pop的node， 如果这个node是新的peek的值的子数的话，不能再push到stack，需要进入输出逻辑。 同样进入输出逻辑的还有leaf。

    public IList<int> PostorderTraversal(TreeNode root) {
        var res = new List<int>();
        var stack = new Stack<TreeNode>();
        if(root == null) return res;
        stack.Push(root);
        var rightPointer = new TreeNode(-1);
        while(stack.Count() > 0)
        {
            if((root.right == null && root.left == null)||(root.left == rightPointer ) || (root.right == rightPointer) )
            {
                res.Add(root.val);
                rightPointer = stack.Pop();
                if(stack.Count()>0)
                root = stack.Peek();
            }
            else
            {
                if(root.right != null) stack.Push(root.right);
                if(root.left != null) stack.Push(root.left);
                root = stack.Peek();
            }

        }
        return res;
    }