Construct Binary Tree from Preorder and Inorder Traversal
題目描述:
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
題目大意:
給定滿二叉樹中序周遊和先序周遊,建立二叉樹。
可以根據先序周遊找出根節點,在中序周遊中找出該節點元素,該節點元素的左邊都是左子樹的節點,右邊都是右子樹的節點。可以通過遞歸建樹。
題目代碼:
/**
* 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) {
return build(preorder, inorder, 0, preorder.size()-1, 0, inorder.size()-1);
}
TreeNode* build(vector<int>& preorder, vector<int>& inorder, int ps, int pe, int is, int ie){
if(ps > pe || is > ie)
return nullptr;
TreeNode* root = new TreeNode(preorder[ps]);
int i = 0;
while(inorder[i] != root->val) i++;
root->left = build(preorder, inorder, ps+1, ps+i-is, is, i-1);
root->right = build(preorder, inorder, ps+i-is+1, pe,i+1, ie);
return root;
}
};