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package leetcode;
import java.util.ArrayList;
import java.util.List;
import java.util.Stack;
/**
* 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?
* @author rwxn
*
*要求使用迭代，不能使用递归，改进后序遍历
*/
public class java_145 {
public class TreeNode {
int val;
TreeNode left;
TreeNode right;
TreeNode(int x) {
val = x;
}
}
public List<Integer> postorderTraversal(TreeNode root) {
List<Integer> list = new ArrayList<>();
Stack<TreeNode> stack = new Stack<TreeNode>();
if (root == null) {
return null;
}
//先将根节点入栈，然后将右子树入栈，再将左子树入栈
stack.push(root);
while (stack.peek() != null) {
TreeNode node = stack.peek();
if (node.right != null) {
stack.push(node.right);
}
if (node.left != null) {
stack.push(node.left);
}
}
while (stack.peek() != null) {
list.add(stack.peek().val);//后序遍历第一个元素出栈并存入list中
stack.pop();
}
return null;
}
}