构造二叉树

构造二叉树

105. 从前序与中序遍历序列构造二叉树

例：

前序遍历 preorder = [3,9,20,15,7]
中序遍历 inorder = [9,3,15,20,7]

二叉树：

	3
   / \
  9  20
    /  \
   15   7

/**
 * 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:
    vector<int> pre, in;
    map<int, int> pos;
    TreeNode* build(int pre_l, int pre_r, int in_l, int in_r){
        if(pre_l > pre_r) return nullptr;
      
        int pre_val = pre[pre_l];
        int p = pos[pre_val];
        
        int num = p-in_l;

        TreeNode* root = new TreeNode(pre_val); 
        
        root-> left = build(pre_l+1, pre_l+num, in_l, p-1);
        root -> right = build(pre_l+num+1, pre_r, p+1, in_r);
        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        pre = preorder; in = inorder;
        for(int i = 0; i < inorder.size(); i++) pos[inorder[i]] = i;
        return build(0, preorder.size()-1, 0, inorder.size()-1);
    }
};

106. 从中序与后序遍历序列构造二叉树

例：

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]

二叉树：

    3
   / \
  9  20
    /  \
   15   7

/**
 * 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:
    vector<int> in, post;
    map<int, int> pos;
    int rt;
    TreeNode* build(int l, int r){
        if(l > r) return nullptr;
      
        int post_val = post[rt];
        int p = pos[post_val];
        
        TreeNode* root = new TreeNode(post_val); 
        rt--;
        root -> right = build(p+1, r);
        root-> left = build(l, p-1);
        
        return root;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        in = inorder; post = postorder;
        rt = inorder.size()-1;
        for(int i = 0; i < inorder.size(); i++) pos[inorder[i]] = i;
        return build(0, rt);
    }
};

889. 根据前序和后序遍历构造二叉树

返回与给定的前序和后序遍历匹配的任何二叉树。

例：

中序遍历 inorder = [9,3,15,20,7]
后序遍历 postorder = [9,15,7,20,3]

二叉树：

      1
     /  \
    2    3
   / \  / \
  4   5 6  7

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode constructFromPrePost(int[] pre, int[] post) {
        int N = pre.length;
        if (N == 0) return null;
        TreeNode root = new TreeNode(pre[0]);
        if (N == 1) return root;

        int L = 0;
        for (int i = 0; i < N; ++i)
            if (post[i] == pre[1])
                L = i+1;

        root.left = constructFromPrePost(Arrays.copyOfRange(pre, 1, L+1),
                                         Arrays.copyOfRange(post, 0, L));
        root.right = constructFromPrePost(Arrays.copyOfRange(pre, L+1, N),
                                          Arrays.copyOfRange(post, L, N-1));
        return root;
    }
}