前序遍歷和中序遍歷樹構造二叉樹
阿新 • • 發佈：2017-06-24
fin traversal dtree 構造二叉樹 div integer break param val
/* */ preorder : A list of integers that preorder traversal of a tree
[email protected] inorder : A list of integers that inorder traversal of a tree
[email protected] : Root of a tree
*/
public:
TreeNode *construct(vector<int> &preorder, vector<int> &inorder, int ps,
int pe, int is, int ie) {
TreeNode * root = new TreeNode(preorder[ps]);
if (ps == pe) return root;
int i;
for (i = 0; i < ie; i++) {
if (inorder[i] == root->val) break; //找到根節點位置
}
//遞歸構造左右子樹
if (is <= i-1)
root->left = construct(preorder, inorder,
ps+1, ps+(i-is), is, i-1);
if (i+1 <= ie)
root->right = construct(preorder, inorder,
ps+(i-is)+1, pe, i+1, ie);
return root;
}
TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
if (preorder.empty() || inorder.empty() ||
preorder.size() != inorder.size()) return NULL;
return construct(preorder, inorder, 0, preorder.size()-1, 0, inorder.size()-1);
}
};
根據前序遍歷和中序遍歷樹構造二叉樹
樣例:
給出中序遍歷：[1,2,3]和前序遍歷：[2,1,3]. 返回如下的樹:
2
/ \
1 3
我們知道前序遍歷是中->左->右，中序遍歷是左->中->右。因此根據前序遍歷的第一個數，即為根節點，我們可以在中序遍歷中找到根節點的左子樹和右子樹，同樣遞歸在左子樹中找到左子樹的根節點和其左右子樹，對右子樹也一樣。這樣理清思路後代碼就不難寫出來了。
/**
* Definition of TreeNode:
* class TreeNode {
* public:
* int val;
* TreeNode *left, *right;
* TreeNode(int val) {
* this->val = val;
* this->left = this->right = NULL;
* }
* }
*/
class Solution {
/**
[email protected]
前序遍歷和中序遍歷樹構造二叉樹