简介
最简单的方法是使用递归来构建整棵树。
最核心的一张图
code
/**
* Definition for a binary tree node.
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode() : val(0), left(nullptr), right(nullptr) {}
* TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
* TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
* };
*/
class Solution {
public:
unordered_map<int, int> index;
public:
TreeNode *myBuildTree(const vector<int>&preorder, const vector<int>&inorder, int preorder_left, int preorder_right
, int inorder_left, int inorder_right)
{
if(preorder_left > preorder_right){
return nullptr;
}
// 前序遍历中的第一个节点就是根节点
int preorder_root = preorder_left;
// 再中序遍历中定位根节点
int inorder_root = index[preorder[preorder_root]];
// 先把根节点建立出来
TreeNode * root = new TreeNode(preorder[preorder_root]);
// 得到左子树中的节点数量
int size_left_subtree= inorder_root - inorder_left;
// 递归的构造左子树, 兵连接到根节点
// 先序遍历中 从左边界 + 1 开始的 size_left_subtree 个元素就对应了中序遍历中, 从左边界开始到根节点的个数
root->left = myBuildTree(preorder, inorder, preorder_left + 1, preorder_left + size_left_subtree, inorder_left, inorder_root - 1);
// 递归地构造右字数, 并连接到根节点
root->right = myBuildTree(preorder, inorder, preorder_left + size_left_subtree + 1, preorder_right, inorder_root + 1, inorder_right);
return root;
}
TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
int n = preorder.size();
for(int i=0; i<preorder.size(); i++){
index[inorder[i]] = i;
}
return myBuildTree(preorder, inorder, 0, n - 1, 0, n - 1);
}
};
/**
* Definition for a binary tree node.
* public class TreeNode {
* int val;
* TreeNode left;
* TreeNode right;
* TreeNode() {}
* TreeNode(int val) { this.val = val; }
* TreeNode(int val, TreeNode left, TreeNode right) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
class Solution {
Map<Integer, Integer> index;
public Solution(){
index = new HashMap<Integer, Integer>();
}
public TreeNode myBuildTree(int [] preorder, int [] inorder, int preorder_left, int preorder_right, int inorder_left, int
inorder_right)
{
if(preorder_left > preorder_right || inorder_left > inorder_right) {
return null;
}
// 前序遍历中的第一个节点就是根节点
int preorder_root = preorder_left;
int inorder_root = index.get(preorder[preorder_root]);
TreeNode root = new TreeNode(preorder[preorder_root]);
int inorder_sub = inorder_root - inorder_left;
root.left = myBuildTree(preorder, inorder, preorder_left + 1, preorder_left + inorder_sub, inorder_left, inorder_root - 1);
root.right = myBuildTree(preorder, inorder, preorder_left + inorder_sub + 1, preorder_right, inorder_root + 1, inorder_right);
return root;
}
public TreeNode buildTree(int[] preorder, int[] inorder) {
for(int i=0; i<preorder.length; i++){
index.put(inorder[i], i);
}
return myBuildTree(preorder, inorder, 0, preorder.length - 1, 0, inorder.length - 1);
}
}