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);

    }

};

巴特西

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

Construct Binary Tree from Preorder and Inorder Traversal

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