给定一个二叉树,返回它的 后序 遍历。
示例:
输入: [1,null,2,3]
1
\
2
/
3
输出: [3,2,1]
进阶: 递归算法很简单,你可以通过迭代算法完成吗?
思路1:递归
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
List<Integer> res =new ArrayList<>();
public List<Integer> postorderTraversal(TreeNode root) {
helper(root);
return res;
}
private void helper(TreeNode root){
if(root==null)
return;
helper(root.left);
helper(root.right);
res.add(root.val);
}
}
思路2:用栈迭代
后序遍历顺序为"左 右 根"
参考前序遍历为"根 左 右" ,可以先转化为"根 右 左",再翻转即可。
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode(int x) { val = x; }
* }
*/
class Solution {
List<Integer> res =new ArrayList<>();
Stack<TreeNode> stack = new Stack<>();
public List<Integer> postorderTraversal(TreeNode root) {
TreeNode curr = root;
while(curr!=null || !stack.isEmpty())
{
while(curr!=null){
stack.push(curr);
res.add(curr.val);
curr = curr.right;
}
curr = stack.pop();
curr = curr.left;
}
Collections.reverse(res);
return res;
}
}