build binary tree from preorder and inorder sequence

Author: wangyuxiang0829

src/105.cpp

@@ -0,0 +1,48 @@
+#include <vector>
+#include <iostream>
+#include <string>
+
+struct TreeNode {
+    int val;
+    TreeNode *left;
+    TreeNode *right;
+    TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
+};
+
+class Solution {
+private:
+    void buildRoot(const std::vector<int> &preorder, const std::vector<int> &inorder, TreeNode *&root) {
+        if (!preorder.size()) return;
+        root = new TreeNode(preorder.front());
+        for (int i = 0; i < inorder.size(); ++i) {
+            if (inorder[i] == preorder.front()) {
+                buildRoot(std::vector<int>(preorder.begin() + 1, preorder.begin() + 1 + i),
+                        std::vector<int>(inorder.begin(), inorder.begin() + i), root->left);
+                buildRoot(std::vector<int>(preorder.begin() + 1 + i, preorder.end()),
+                        std::vector<int>(inorder.begin() + i + 1, inorder.end()), root->right);
+                return;
+            }
+        }
+    }
+
+public:
+    TreeNode *buildTree(const std::vector<int> &preorder, const std::vector<int> &inorder) {
+        TreeNode *root = nullptr;
+        buildRoot(preorder, inorder, root);
+        return root;
+    }
+};
+
+void traversing(TreeNode *root, unsigned level) {
+    if (!root) return;
+    traversing(root->right, level + 1);
+    std::cout << std::string(4 * level, ' ') << root->val << std::endl;
+    traversing(root->left, level + 1);
+}
+
+int main() {
+    Solution s;
+    traversing(s.buildTree(std::vector<int>({3, 9, 20, 15, 7}), std::vector<int>({9, 3, 15, 20, 7})), 0);
+    return 0;
+}
+