思路:根據先序和後序遍歷數據,遞歸的構造整棵樹
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> &postorder) {
buildPos(postorder);
return buildTree(preorder, 0, preorder.size()-1, postorder, 0, postorder.size()-1);
}
TreeNode *buildTree(vector<int> &pre, int l, int r, vector<int> &post, int b, int e) {
if(l > r || b > e) return NULL;
TreeNode *root = new TreeNode(pre[l]);
if(l < r)
{
int postIdx = pos[pre[l+1]], preIdex = postIdx + l + 1;
root->left = buildTree(pre, l+1, preIdex, post, b, postIdx);
root->right = buildTree(pre, preIdex+1, r, post, postIdx+1, e-1);
}
return root;
}
void buildPos(vector<int> &postorder) {
for(int i = 0; i < postorder.size(); ++i)
pos[postorder[i]] = i;
}
private:
unordered_map<int, int> pos;
};