Construct Binary Tree from Inorder and Postorder Traversal

Description

Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.

Example 1:

Input: inorder = [9,3,15,20,7], postorder = [9,15,7,20,3] Output: [3,9,20,null,null,15,7]

Example 2:

Input: inorder = [-1], postorder = [-1] Output: [-1]

Constraints:

• 1 <= inorder.length <= 3000
• postorder.length == inorder.length
• -3000 <= inorder[i], postorder[i] <= 3000
• inorder and postorder consist of unique values.
• Each value of postorder also appears in inorder.
• inorder is guaranteed to be the inorder traversal of the tree.
• postorder is guaranteed to be the postorder traversal of the tree.

Code

同 Construct Binary Tree from Preorder and Inorder Traversal。

Trick：postorder （LRN → NRL）反過來就類似 preorder（NLR）。這技巧在 N-ary Tree Postorder Traversal 中用過。

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode() : val(0), left(nullptr), right(nullptr) {}
 *     TreeNode(int x) : val(x), left(nullptr), right(nullptr) {}
 *     TreeNode(int x, TreeNode *left, TreeNode *right) : val(x), left(left), right(right) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        int rootIdx = 0;
        // LRN -> NRL, simliar to preorder
        reverse(postorder.begin(), postorder.end());
        return build(postorder, inorder, rootIdx, 0, inorder.size()-1);
    }
    
    TreeNode* build(vector<int>& postorder, vector<int>& inorder, int& rootIdx, int left, int right) {
        if (left > right) return NULL;
        int pivot = left;  // find the root from inorder
        while(inorder[pivot] != postorder[rootIdx]) pivot++;
        
        rootIdx++;
        TreeNode* newNode = new TreeNode(inorder[pivot]);
        newNode->right = build(postorder, inorder, rootIdx, pivot+1, right);
        newNode->left = build(postorder, inorder, rootIdx, left, pivot-1);
        return newNode;
    }
};

Source

• Construct Binary Tree from Inorder and Postorder Traversal - LeetCode