145. Binary Tree Postorder Traversal

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]

.

将后序遍历的结果逆序，就是先序遍历时先遍历右子树，再遍历左子树的结果。因此只需在先序遍历的结果上稍微改变一下代码。

public class Solution {
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> list=new ArrayList<>();
Stack<TreeNode> stack=new Stack<>();
TreeNode p=root;
while(p!=null||!stack.isEmpty()){
while(p!=null){
list.add(0,p.val);
stack.push(p);
p=p.right;
}
if(!stack.isEmpty()){
p=stack.pop();
p=p.left;
}
}
return list;
}
}