leetcode 144. Binary Tree Preorder Traversal

原文链接http://www.myexception.cn/program/1958647.html

题目

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

For example:

Given binary tree {1,#,2,3},



return [1,2,3].

解1（递归）

//前序遍历     递归
public static void preorderTraversal(TreeNode root){

if(root==null) return ;

System.out.print(root.val+"  ");
inorderTraversal(root.left);
inorderTraversal(root.right);
}

解2（非递归方法）

/**
* Definition for a binary tree node.
* 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> res=new ArrayList<Integer>();
Stack<TreeNode> nodeStack=new Stack<>();

while(true)
{
while(root!=null)
{
res.add(root.val);
nodeStack.push(root);
root=root.left;
}
if(nodeStack.isEmpty()) break;

TreeNode tempNode=nodeStack.pop();
root=tempNode.right;
}
return res;
}
}

解3（非递归）

//前序遍历    非递归
public static List<TreeNode> preorderTraversal2(TreeNode root){

List<TreeNode> tList=new ArrayList<TreeNode>();
Stack<TreeNode> tStack=new Stack<TreeNode>();

tStack.push(root);
while(!tStack.isEmpty()){
TreeNode p=tStack.pop();
tList.add(p);
System.out.print(p.val+" ");

if(p.right!=null) tStack.push(p.right);
if(p.left!=null)  tStack.push(p.left);
}

return tList;

}