LeetCode106. Construct Binary Tree from Inorder and Postorder Traversal

106. 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.

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

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题目：根据给定的中序和后序排列，重建二叉树。

思路：思路同LeetCode105. Construct Binary Tree from Preorder and Inorder Traversal。注意判断postorder对应左子树的结束位置和右子树的起始位置。

工程代码下载

/**
 * 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:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return buildTree(0, inorder.size()-1, 0, postorder.size()-1, inorder, postorder);
    }
private:
    TreeNode* buildTree(int in_start, int in_end, int post_start, int post_end,
                       vector<int>& inorder, vector<int>& postorder){
        if(in_start > in_end || post_start > post_end){
            return nullptr;
        }

        TreeNode* root = new TreeNode(postorder[post_end]);

        int in_root_index = 0;
        for(int i = in_start; i <= in_end; ++i){
            if(inorder[i] == root->val){
                in_root_index = i;
                break;
            }
        }

        root->left = buildTree(in_start, in_root_index - 1,
                               post_start, post_start + in_root_index - in_start - 1,
                              inorder, postorder);
        root->right = buildTree(in_root_index + 1, in_end,
                               post_start + in_root_index - in_start, post_end - 1,
                               inorder, postorder);
        return root;
    }
};