Given a binary tree, return the postorder traversal of its nodes‘ values.
For example:
Given binary tree {1,#,2,3},
1
2
/
3
return [3,2,1].
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> postorderTraversal(TreeNode root) {
List<Integer> list=new ArrayList<Integer>();
post(root,list);
return list;
}
public void post(TreeNode node,List<Integer> list){
if(node==null) return;
post(node.left,list);
post(node.right,list);
list.add(node.val);
}
}思路:二叉树的递归后序遍历,非递归的也得会。Binary Tree Postorder Traversal,布布扣,bubuko.com
Binary Tree Postorder Traversal
原文地址:http://blog.csdn.net/dutsoft/article/details/37737725