Binary Tree Postorder Traversal

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?

思路：

（1）題意為後序遍歷二叉樹。遍歷順序為左—>右—>根。

（2）考慮到用遞歸比較簡單，本文使用遞歸的思想進行解決，由於比較簡單這裡不累贅，詳見下方代碼。

（3）希望本文對你有所幫助。

算法代碼實現如下：

/**
 * @author liqq
 */
public List PostorderTraversal(TreeNode root) {
	List result = new LinkedList();
	if (root != null) {
		Post_order(result, root.left);
		Post_order(result, root.right);
		result.add(root.val);
	}
	return result;
}

private void Post_order(List result, TreeNode curr) {
	if (curr != null) {
		Post_order(result, curr.left);
		Post_order(result, curr.right);
		result.add(curr.val);
	}
}