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;
     }
 }