关关的刷题日记98 – Leetcode 106. Construct Binary Tree from Inorder- 专知

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关关的刷题日记98 – Leetcode 106. Construct Binary Tree from Inorder

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关关的刷题日记98 – Leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal

题目

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

思路

给出二叉树的后根遍历和中根遍历，构造二叉树。

了解后根遍历和中根遍历的特点。

每次在后根遍历数组中最右边就是二叉树的根，然后去中根遍历数组中找到根，根左边的就是左子树，根右边的就是右子树，对于左子树和右子树采取同样的求法：去后根遍历数组中找根，利用中根遍历数组划分左右子树，递归求解。

class Solution {
public:
    TreeNode * help(vector < int > & postorder, int


l1, int
r1, vector < int > & inorder, int
l2, int
r2)
{
if (l1 > r1)
return nullptr;
else
{
    int
i;
TreeNode * root = new
TreeNode(postorder[r1]);
for (i=l2; i <= r2; ++i)
{
if (inorder[i] == postorder[r1])
break;
}
root->left = help(postorder, l1, l1 + i - 1 - l2, inorder, l2, i - 1);
root->right = help(postorder, l1 + i - l2, r1 - 1, inorder, i + 1, r2);
return root;
}
}

TreeNode * buildTree(vector < int > & inorder, vector < int > & postorder) {
return help(postorder, 0, postorder.size() - 1, inorder, 0, inorder.size() - 1);
}
};