关关的刷题日记97 – Leetcode 105. Construct Binary Tree- 专知

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关关的刷题日记97 – Leetcode 105. Construct Binary Tree

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关关的刷题日记97 – Leetcode 105. Construct Binary Tree

题目

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

Note:

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

思路

给出二叉树的先根遍历和中根遍历，构造二叉树。

了解先根遍历和中根遍历的特点。每次在先根遍历数组中最左边的就是二叉树的根，然后去中根遍历数组中找到根，根左边的就是左子树，对于左子树和右子树采取同样的求法：去先根遍历数组中找根，利用中根遍历数组划分左右子树，递归求解。

/ **
*Definition
for a binary tree node.
*struct
TreeNode
{
    *int
val;
*TreeNode * left;
*TreeNode * right;
*TreeNode(int
x): val(x), left(NULL), right(NULL)
{}
*};
* /

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


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

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