106. 从中序与后序遍历序列构造二叉树
题目链接
来源:力扣(LeetCode)
链接:https://leetcode-cn.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/
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题目描述
根据一棵树的中序遍历与后序遍历构造二叉树。
注意:
你可以假设树中没有重复的元素。
例如,给出
中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]
返回如下的二叉树:
3
/ \
9 20
/ \
15 7
题目分析
- 根据后序和中序构造二叉树
- 根据后序序列找到父节点,再根据父节点在中序序列中找出该父节点的左右子树
- 按照步骤2级进行递归构造结点
代码
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
int inorderLen = inorder.size();
int postorderLen = postorder.size();
if (inorderLen == 0 || postorderLen == 0) return nullptr;
return recursion(inorder, 0, inorderLen - 1, postorder, 0, postorderLen - 1);
}
private:
TreeNode* recursion(vector<int>& inorder, int inorderLeft, int inorderRight, vector<int>& postorder, int postorderLeft, int postorderRight) {
if (inorderLeft > inorderRight || postorderLeft > postorderRight) return nullptr;
TreeNode* node = new TreeNode(postorder[postorderRight]);
int mid;
for (mid = inorderLeft; mid < inorderRight; ++ mid) {
if (inorder[mid] == postorder[postorderRight]) break;
}
node->left = recursion(inorder, inorderLeft, mid - 1, postorder, postorderLeft, postorderLeft + mid - inorderLeft - 1);
node->right = recursion(inorder, mid + 1, inorderRight, postorder, postorderLeft + mid - inorderLeft, postorderRight - 1);
return node;
}
};