LeetCode 144 Binary Tree Preorder Traversal (先序遍历二叉树)

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?

题目链接：https://leetcode.com/problems/binary-tree-preorder-traversal/

题目分析：

递归

/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
public class Solution {
void DFS(TreeNode root, List<Integer> ans) {
if(root == null) {
return;
}
ans.add(root.val);
DFS(root.left, ans);
DFS(root.right, ans);
}

public List<Integer> preorderTraversal(TreeNode root) {
List<Integer> ans = new ArrayList<>();
DFS(root, ans);
return ans;
}
}
非递归的方法就是拿栈模拟，中往左走的时候直接遍历，走到不能走然后再往右

/**
* 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> ans = new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
while(root != null || stack.size() > 0) {
while(root != null) {
ans.add(root.val);
stack.push(root);
root = root.left;
}
if(stack.size() > 0) {
root = stack.peek();
stack.pop();
root = root.right;
}
}
return ans;
}
}