【二叉树构造】654. Maximum Binary Tree; 105. Construct Binary Tree from Preorder and Inorder Traversal; 106. Construct Binary Tree from Inorder and Postorder Traversal

654. Maximum Binary Tree

问题：

用给定的数组，构成最大二叉树。

从数组中找到最大值，作为root，其index以左，作为左子树。index以右，作为右子树。

Example 1:
Input: [3,2,1,6,0,5]
Output: return the tree root node representing the following tree:

      6
    /   \
   3     5
    \    / 
     2  0   
       \
        1
Note:
The size of the given array will be in the range [1,1000].

　　

解法：Binary Tree（二叉树）

递归：help函数

• base：数组为空：end>=start
  • 返回 nullptr
• 对于每个节点root ：在nums[start, end)范围内
  • 找到最大值maxval，及其index：idx
    • root->val = maxval
    • root->left = 递归求解：nums[start, idx]　
    • root->right = 递归求解：nums[idx+1, end]

代码参考：

/**
 * 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:
    TreeNode* help(vector<int>& nums, int start, int end) {
        if(end<=start) return nullptr;
        int maxval = INT_MIN;
        int idx = 0;
        for(int i=start; i<end; i++) {
            if(nums[i]>maxval) {
                maxval = nums[i];
                idx = i;
            }
        }
        TreeNode* root = new TreeNode(maxval);
        //child:
        root->left = help(nums, start, idx);
        root->right = help(nums, idx+1, end);
        return root;
    }
    TreeNode* constructMaximumBinaryTree(vector<int>& nums) {
        return help(nums, 0, nums.size());
    }
};

 

105. Construct Binary Tree from Preorder and Inorder Traversal

labuladong 参考说明

问题：

由【前序遍历】和【中序遍历】数组，构造二叉树。

For example, given

preorder = [3,9,20,15,7]
inorder = [9,3,15,20,7]
Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

　　

解法：Binary Tree（二叉树）

递归：help函数

• base：数组为空：preE>=preS or inE>=inS
  • 返回 nullptr
• 对于每个节点root ：在preorder[preS, preE)和inorder[inS, inE)范围内
  • 在中序中，找到前序遍历的第一个值val：preorder[preS]，的index：idx。求出左树的长度：leftSize = idx-inS+1;
    • root->val = preorder[preS]
    • root->left = 递归求解：preorder[preS+1, preS+leftSize) 和 inorder[inS, idx)
    • root->right = 递归求解：preorder[preS+leftSize, preE) 和 inorder[idx+1, inE)

 

 

 

代码参考：

/**
 * 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:
    //preorder[preS, preE)
    //inorder[inS, inE)
    TreeNode* help(vector<int>& preorder, int preS, int preE,
                   vector<int>& inorder, int inS, int inE) {
        if(preS >= preE || inS >= inE) return nullptr;
        int val=preorder[preS], idx=0;
        for(int i=inS; i<inE; i++) {
            if(inorder[i] == val) {
                idx = i;
                break;
            }
        }
        //root
        TreeNode *root = new TreeNode(val);
        //child
        int leftSize = idx-inS+1;
        root->left = help(preorder, preS+1, preS+leftSize,
                         inorder, inS, idx);
        root->right = help(preorder, preS+leftSize, preE,
                         inorder, idx+1, inE);
        return root;
    }
    TreeNode* buildTree(vector<int>& preorder, vector<int>& inorder) {
        return help(preorder, 0, preorder.size(),
                   inorder, 0, inorder.size());
    }
};

 

106. Construct Binary Tree from Inorder and Postorder Traversal

labuladong 参考说明

问题：

由【后序遍历】和【中序遍历】数组，构造二叉树。

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:

    3
   / \
  9  20
    /  \
   15   7

　　

解法：Binary Tree（二叉树）

递归：help函数

• base：数组为空：postE>=postS or inE>=inS
  • 返回 nullptr
• 对于每个节点root ：在postorder[postS, postE)和inorder[inS, inE)范围内
  • 在中序中，找到后序遍历的最后一个值val：postorder[postE-1]，的index：idx。求出左树的长度：leftSize = idx-inS;
    • root->val = postorder[postE-1]
    • root->left = 递归求解：postorder[postS, postS+leftSize) 和 inorder[inS, idx)
    • root->right = 递归求解：postorder[postS+leftSize, postE-1) 和 inorder[idx+1, inE)

 

 

代码参考：

/**
 * 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:
    TreeNode* help(vector<int>& inorder, int inS, int inE,
                   vector<int>& postorder, int postS, int postE) {
        if(inE<=inS || postE<=postS) return nullptr;
        int val = postorder[postE-1], idx=0;
        for(int i=inS; i<inE; i++) {
            if(inorder[i] == val) {
                idx = i;
                break;
            }
        }
        //root
        TreeNode *root = new TreeNode(val);
        int leftSize = idx-inS;
        //child
        root->left = help(inorder, inS, idx,
                         postorder, postS, postS+leftSize);
        root->right = help(inorder, idx+1, inE,
                         postorder, postS+leftSize, postE-1);
        return root;
    }
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return help(inorder, 0, inorder.size(),
                   postorder, 0, postorder.size());
    }
};