106. Construct Binary Tree from Inorder and Postorder Traversal

1.問題

2.想法

• 提問

  • 確認題意: root是NULL是true或false?

• function header, parameter

• r input

• 說明想法

  • 想法與105題相同, 由於postorder序列建構tree, 必須由序列的最後往前

• 測試計算複雜度

3.程式碼

/**
 * 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) {
        int pos = postorder.size() - 1;
        return buildBinary(inorder, postorder, 0, postorder.size() - 1, pos);
    }
private:
    TreeNode* buildBinary(vector<int>& inorder, vector<int>& postorder, int left, int right, int& pos) {
        if (pos < 0 || left > right) {
            return NULL;
        }
        int i = 0;
        for (i = left; i <= right; i++) {
            if (inorder[i] == postorder[pos]){
                break;
            }
        }
        TreeNode* node = new TreeNode(postorder[pos]);
        pos--;
        node->right = buildBinary(inorder, postorder, i + 1, right, pos);
        node->left = buildBinary(inorder, postorder, left, i - 1, pos);
        
        return node;
    }
};