标签:binary tree preorder traversal
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?
/**
* 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;
}
}leetcode_144_Binary Tree Preorder Traversal
标签:binary tree preorder traversal
原文地址:http://blog.csdn.net/mnmlist/article/details/44491599