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

For example:

Given binary tree {1,#,2,3},

1    \     2    /   3

 

return [1,2,3].

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

[解题思路]

递归解法

1 public ArrayList
     
       preorderTraversal(TreeNode root) { 2         // IMPORTANT: Please reset any member data you declared, as 3         // the same Solution instance will be reused for each test case. 4         ArrayList
      
        result = new ArrayList
       
        (); 5          6         preorder(root, result); 7          8         return result; 9     }10     11     public void preorder(TreeNode root, ArrayList
        
          result){12         if(root != null){13             result.add(root.val);14             preorder(root.left, result);15             preorder(root.right, result);16         }17     }
        
       
      
     

[非递归解法]

1)访问节点cur，并将节点cur入栈；

2)判断节点cur的左孩子是否为空，若为空，则取栈顶节点并进行出栈操作，并将栈顶节点的右孩子置为当前节点cur，循环至1);若不为空，则将cur的左孩子置为当前节点cur；

3)知道cur为null并且栈为空，则遍历结束

1 public ArrayList
     
       preorderTraversal(TreeNode root) { 2         // IMPORTANT: Please reset any member data you declared, as 3         // the same Solution instance will be reused for each test case. 4         ArrayList
      
        result = new ArrayList
       
        (); 5         if(root == null){ 6             return result; 7         } 8          9         Stack
        
          stack = new Stack
         
          ();10         TreeNode cur = root;11         while(cur != null || !stack.empty()){12             13             while(cur != null){14                 result.add(cur.val);15                 stack.push(cur);16                 cur = cur.left;17             }18             19             if(!stack.empty()){20                 cur = stack.pop();21                 // if(cur.right != null){
    22                     cur = cur.right;23                 // }24             }25         }26         27         return result;28     }