Java [Leetcode 144]binary Tree Preorder traversal

Title Description:

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] .

Problem Solving Ideas:

The simplest way to do this is to use recursion, but the problem does not apply, and the iterative approach is used.

Then we'll consider using stacks to implement.

The idea is to go through this node each time, put the value of the point into the list, and then put the right child of the point into the stack, and set the current point to the left child, and then iterate to access the current point, if the left child encountered the current point is empty, then pop up the point in the stack (if the stack is not empty)

The code is as follows:

/** * 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> stack = new stack<treenode> ();    TreeNode node = root;    while (node! = null) {    res.add (node.val);    if (node.right! = null)    Stack.push (node.right);    node = node.left;    if (node = = null &&!stack.isempty ())    node = Stack.pop ();    }    return res;    }}

　　

Java [Leetcode 144]binary Tree Preorder traversal