leetcode_144_Binary Tree Preorder Traversal

描述：

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?

思路：

具体思路和中序遍历是一致的，只是访问结点的值的时机不同罢了，具体思路参见：http://blog.csdn.net/mnmlist/article/details/44312315

代码：

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public List<Integer> preorderTraversal(TreeNode root) {
	     List<Integer>list=new ArrayList<Integer>();
	     if(root==null)
			 return list;
	     Stack<TreeNode>st=new Stack<TreeNode>();
	     st.push(root);
	     TreeNode top=null;
	     list.add(root.val);
	     while(!st.empty())
	     {
	    	 top=st.peek();
	    	 while(top.left!=null)
	    	 {
	    		 st.push(top.left);
	    		 list.add(top.left.val);
	    		 top=top.left;
	    	 }
	    	 while(top.right==null)
	    	 {

	    		 st.pop();
	    		 if(!st.empty())
	    			 top=st.peek();
	    		 else
	    			 break;
	    	 }
	    	 if(!st.empty())
	    	 {
		    	 st.pop();
		    	 st.push(top.right);
		    	 list.add(top.right.val);
	    	 }

	     }
	     return list;
	 }
}

结果：