36. Construct Binary Tree from Inorder and Postorder Traversal && Construct Binary Tree from Preorder and Inorder Traversal

Construct Binary Tree from Inorder and Postorder Traversal

OJ:　https://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

Given inorder and postorder traversal of a tree, construct the binary tree.

Note: You may assume that duplicates do not exist in the tree.

思想： 迭代。

说明： 这类问题，要求一个提供根节点，然后另一个序列（中序序列）可依据根节点分出左右子树。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
TreeNode *createTree(vector<int> &inorder, int start, int end, vector<int> &postorder, int start2, int end2) {
    if(start > end || start2 > end2) return NULL;
    TreeNode *root = new TreeNode(postorder[end2]);
    int i;
    for(i = start; i <= end; ++i) 
        if(inorder[i] == postorder[end2]) break;
   // if(i > end) throws std::exception("error");
    root->left = createTree(inorder, start, i-1, postorder, start2, start2 + i-start-1);
    root->right = createTree(inorder, i+1, end, postorder, start2+i-start, end2-1);
    return root;
}
 
class Solution {
public:
    TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder) {
        return createTree(inorder, 0, inorder.size()-1, postorder, 0, postorder.size()-1);
    }
};

 

Construct Binary Tree from Preorder and Inorder Traversal

 

Given preorder and inorder traversal of a tree, construct the binary tree.

Note: You may assume that duplicates do not exist in the tree.

思想： 同上。

/**
 * Definition for binary tree
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
TreeNode* createTree(vector<int> &preorder, int start, int end, vector<int> &inorder, int start2, int end2) {
    if(start > end || start2 > end2) return NULL;
    TreeNode *root = new TreeNode(preorder[start]);
    int i;
    for(i = start2; i <= end2; ++i) 
        if(preorder[start] == inorder[i]) break;
    root->left = createTree(preorder, start+1, start+i-start2, inorder, start2, i-1);
    root->right = createTree(preorder, start+i-start2+1, end, inorder, i+1, end2);
    return root;
} 
class Solution {
public:
    TreeNode *buildTree(vector<int> &preorder, vector<int> &inorder) {
        return createTree(preorder, 0, preorder.size()-1, inorder, 0, inorder.size()-1);
    }
};