LeetCode 0106. Construct Binary Tree from Inorder and Postorder Traversal【二叉树】

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

---

根据后序序列和中序序列构建二叉树

思路1

代码1

/**
 * 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 build(inorder, postorder, 0, postorder.size()-1, 0, postorder.size()-1);
    }
    
    TreeNode* build(vector<int>& inorder, vector<int>& postorder, int pl, int pr, int il, int ir){
        if(pl > pr) return NULL;
        TreeNode* node = new TreeNode(postorder[pr]);
        int k = 0;
        for(int i = il; i <= ir; i++)
            if(inorder[i] == node->val)
            {
                k = i;
                break;
            }
        node->left = build(inorder, postorder, pl, pl + k - il - 1, il, k - 1);
        node->right = build(inorder, postorder, pl + k -il, pr - 1, k + 1, ir);
        return node;
    }
};