LC145. 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].
Note: Recursive solution is trivial, could you do it iteratively?
啊这题这不是Morries Traversal 么… 见LC 99
Stack solution：
利用postorder是preorder的倒序。先将pre改成中右左，再reverse.
/**
* 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> postorderTraversal(TreeNode root) {
LinkedList<Integer> res = new LinkedList<>();
Deque<TreeNode> stack = new ArrayDeque<>();
if(root == null) return res;
TreeNode p = root;
while(!stack.isEmpty() || p != null){
if(p != null){
stack.push(p);
res.addFirst(p.val);
p = p.right;
}else{
TreeNode n = stack.pop();
p = n.left;
}
} // while
return res;
}
}