solve #145

Author: aylei

src/lib.rs

@@ -139,3 +139,4 @@ mod n0139_word_break;
 mod n0140_word_break_ii;
 mod n0143_reorder_list;
 mod n0144_binary_tree_preorder_traversal;
+mod n0145_binary_tree_postorder_traversal;

src/n0145_binary_tree_postorder_traversal.rs

@@ -0,0 +1,55 @@
+/**
+ * [145] Binary Tree Postorder Traversal
+ *
+ * Given a binary tree, return the postorder traversal of its nodes' values.
+ * 
+ * Example:
+ * 
+ * 
+ * Input: [1,null,2,3]
+ *    1
+ *     \
+ *      2
+ *     /
+ *    3
+ * 
+ * Output: [3,2,1]
+ * 
+ * 
+ * Follow up: Recursive solution is trivial, could you do it iteratively?
+ * 
+ */
+pub struct Solution {}
+use super::util::tree::{TreeNode, to_tree};
+
+// submission codes start here
+
+use std::rc::Rc;
+use std::cell::RefCell;
+impl Solution {
+    pub fn postorder_traversal(root: Option<Rc<RefCell<TreeNode>>>) -> Vec<i32> {
+        let mut res = Vec::new();
+        Solution::helper(root, &mut res);
+        res
+    }
+
+    fn helper(root: Option<Rc<RefCell<TreeNode>>>, vec: &mut Vec<i32>) {
+        if let Some(node) = root {
+            Solution::helper(node.borrow().left.clone(), vec);
+            Solution::helper(node.borrow().right.clone(), vec);
+            vec.push(node.borrow().val);
+        }
+    }
+}
+
+// submission codes end
+
+#[cfg(test)]
+mod tests {
+    use super::*;
+
+    #[test]
+    fn test_145() {
+        assert_eq!(Solution::postorder_traversal(tree![1,null,2,3]), vec![3,2,1]);
+    }
+}