106. Construct Binary Tree from Inorder and Postorder Traversal

难度: Medium

题目描述

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.

解题思路

代码实现

解决方案

java

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        
        Map<Integer,Integer> inMap  = new HashMap<>();
        for(int i=0;i<inorder.length;i++){
            inMap.put(inorder[i],i);
        }

        return builder(postorder,inorder,0,postorder.length-1,0,inorder.length-1,inMap);

    }

    public TreeNode builder(int[] postorder, int[] inorder, int preLeft, int preRight, int inLeft, int inRight,
            Map<Integer, Integer> inMap) {

        if (preLeft > preRight) {
            return null;
        }
        TreeNode root = new TreeNode(postorder[preRight]);

        int rootInOrderIndex = inMap.get(postorder[preRight]);
        int leftLength = rootInOrderIndex - inLeft;
        int rightLength = inRight - rootInOrderIndex;

        TreeNode left = builder(postorder, inorder, preLeft, preLeft + leftLength-1, inLeft, rootInOrderIndex - 1,
                inMap);
        TreeNode right = builder(postorder, inorder, preLeft + leftLength, preRight-1, rootInOrderIndex + 1, inRight,
                inMap);
        root.left = left;
        root.right = right;
        return root;

    }
}