build tree

Author: hxzhao527

datastructure/src/tree.rs

@@ -10,6 +10,17 @@ pub struct TreeNode {
     pub right: Option<Rc<RefCell<TreeNode>>>,
 }
 
+impl TreeNode {
+  #[inline]
+  pub fn new(val: i32) -> Self {
+    TreeNode {
+      val,
+      left: None,
+      right: None
+    }
+  }
+}
+
 impl PartialEq for TreeNode {
     fn eq(&self, other: &Self) -> bool {
         if self.val != other.val {

src/tree/traversal/mod.rs

@@ -3,3 +3,4 @@
 pub mod in_order;
 pub mod post_order;
 pub mod pre_order;
+pub mod no_class;

src/tree/traversal/no_class.rs

@@ -0,0 +1,50 @@
+use datastructure::TreeNode;
+use std::cell::RefCell;
+use std::rc::Rc;
+
+/// [105. 从前序与中序遍历序列构造二叉树](https://leetcode.cn/problems/construct-binary-tree-from-preorder-and-inorder-traversal/)
+///
+/// ## 思路
+/// 1. 通过前序知道根节点，然后分割中序得到左右子树
+///    - 通过中序找到根节点的位置, 然后通过左右子树长度分割前序
+/// 2. 递归构建左右子树
+///     - 递归结束条件是前序为空
+/// 3. 挂载左右子树到根, 然后返回根节点
+pub fn build_tree(preorder: Vec<i32>, inorder: Vec<i32>) -> Option<Rc<RefCell<TreeNode>>> {
+    fn build(preorder: &[i32], inorder: &[i32]) -> Option<Rc<RefCell<TreeNode>>> {
+        if preorder.is_empty() {
+            return None;
+        }
+        let root_val = preorder[0];
+        let mut root = TreeNode::new(root_val);
+
+        let root_idx = inorder.iter().position(|&x| x == root_val).unwrap();
+        let (inorder_left, inorder_right) = inorder.split_at(root_idx);
+
+        let inorder_left_len = inorder_left.len();
+        let (preorder_left, preorder_right) = preorder.split_at(inorder_left_len + 1);
+
+        root.left = build(&preorder_left[1..], inorder_left);
+        root.right = build(preorder_right, &inorder_right[1..]);
+        Some(Rc::new(RefCell::new(root)))
+    }
+    build(&preorder, &inorder)
+}
+
+#[cfg(test)]
+mod tests {
+    use macros::tree;
+
+    use super::*;
+
+    #[test]
+    fn test_build_tree() {
+        let preorder = vec![3, 9, 20, 15, 7];
+        let inorder = vec![9, 3, 15, 20, 7];
+        let root = build_tree(preorder, inorder);
+        assert_eq!(
+            root,
+            tree!({3, left: {9}, right:{20, left: {15}, right:{7} }})
+        );
+    }
+}