144. Binary Tree Preorder Traversal (二叉树前序遍历)

---

Given a binary tree, return the preorder traversal of its nodes' values.

For example:
Given binary tree [1,null,2,3],

   1
    
     2
    /
   3

return [1,2,3].

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

递归 ：

class Solution {
    private List<Integer> res = new ArrayList<Integer>();
    public List<Integer> preorderTraversal(TreeNode root) {
        help(root);
        return res;
    }
    private void help(TreeNode root){
        if(root == null) return ;
        res.add(root.val);
        help(root.left);
        help(root.right);
    }
}

非递归：

终极版：

class Solution {
    public List<Integer> preorderTraversal(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(root.val);
                root = root.left;
            }
            if(!s.isEmpty()){
                root = s.pop();
                root = root.right;
            }
        }
        return res;
    }
}

class Solution {   
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        
        //用stack 来保存 右子树
        Stack<TreeNode> stack = new Stack<TreeNode>();
        TreeNode cur = root;
        while(cur!=null || !stack.isEmpty()){
            while(cur!=null){
                res.add(cur.val);
                stack.add(cur.right);
                cur = cur.left;
            }
            cur = stack.pop();
        }
        return res;
    }   
}

 20180320

用stack 保存右节点跟左节点

class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
        List<Integer> res = new ArrayList<Integer>();
        Stack<TreeNode> stack = new Stack();
        stack.push(root);
        while(!stack.isEmpty()){
            root = stack.pop();
            if (root!=null){
            res.add(root.val);
            stack.push(root.right);
            stack.push(root.left);
            }
        }
        return res;
    }
}