105. Construct Binary Tree from Preorder and Inorder Traversal(构建二叉树) [LeetCode]

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本文介绍如何从先序遍历与中序遍历、中序遍历与后序遍历构造二叉树的方法，并提供了详细的代码实现。通过递归的方式确定根节点及其左右子树，最终构建出完整的二叉树。

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105. Construct Binary Tree from Preorder and Inorder Traversal

For example, given：

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]

Return the following binary tree:

画了一个复杂点的二叉树来剖析本题的递归，注意尾递归不在搜索范围内，只是用来确定边界，边界一定要考虑清除！

class Solution {
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        return buildTree(preorder.data(), inorder.data(), inorder.size());
    }

private:
    TreeNode *buildTree(int *preorder, int *inorder, int size) {
        if (NULL == preorder || NULL == inorder || size <= 0) return NULL;

        TreeNode *root = new TreeNode(preorder[0]);
        if (size == 1)  //recursion terminate, trace back
            return root;

        int rootIndex = 0;  //search the root index in inorder
        for (; rootIndex < size; rootIndex++)
            if (inorder[rootIndex] == root->val) break;

        root->left = buildTree(&preorder[1], inorder, rootIndex);
        root->right = buildTree(&preorder[rootIndex + 1], &inorder[rootIndex + 1], size - rootIndex - 1);

        return root;
    }
};

106. Construct Binary Tree from Inorder and Postorder Traversal

given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree:

同样以上面的图为例来说明：

class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return buildTree(postorder.data(), inorder.data(), inorder.size());
    }
private:
    TreeNode *buildTree(int *postorder, int *inorder, int size) {
        if (NULL == postorder || NULL == inorder || size <= 0) return NULL;

        TreeNode *root = new TreeNode(postorder[size-1]);
        if (size == 1)  //recursion terminate, trace back
            return root;

        int rootIndex = 0;  //search the root index in inorder
        for (; rootIndex < size; rootIndex++)
            if (inorder[rootIndex] == root->val) break;

        root->left = buildTree(&postorder[0], inorder, rootIndex);
        root->right = buildTree(&postorder[rootIndex], &inorder[rootIndex + 1], size - rootIndex - 1);

        return root;
    }
};