tag:
- Medium；
question
Given preorder and inorder traversal of a tree, construct the binary tree.
Note: You may assume that duplicates do not exist in the tree.
Example:
preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:
3
/ \
9 20
/ \
15 7
思路：
由于先序的顺序的第一个肯定是根，所以原二叉树的根节点可以知道，题目中给了一个很关键的条件就是树中没有相同元素，有了这个条件我们就可以在中序遍历中也定位出根节点的位置，并以根节点的位置将中序遍历拆分为左右两个部分，分别对其递归调用原函数即可。代码如下：
/**
* 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* buildTree(vector<int>& preorder, vector<int>& inorder) {
if (preorder.size() != inorder.size()) return NULL;
return buildTree(preorder, 0, preorder.size() - 1, inorder, 0, inorder.size() - 1);
}
TreeNode* buildTree(vector<int>& preorder, int preLeft, int preRight, vector<int>& inorder, int inLeft, int inRight) {
if (preLeft > preRight || inLeft > inRight) return NULL;
int i = 0;
for (i=inLeft; i<inRight; ++i) {
if (preorder[preLeft] == inorder[i]) break;
}
TreeNode *cur = new TreeNode(preorder[preLeft]);
cur->left = buildTree(preorder, preLeft + 1, preLeft+i-inLeft, inorder, inLeft, i-1);
cur->right = buildTree(preorder, preLeft+i-inLeft+1, preRight, inorder, i+1, inRight);
return cur;
}
};
