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  • [LC] 105. Construct Binary Tree from Preorder and Inorder Traversal

    Given preorder and inorder traversal of a tree, construct the binary tree.

    Note:
    You may assume that duplicates do not exist in the tree.

    For example, given

    preorder = [3,9,20,15,7]
    inorder = [9,3,15,20,7]

    Return the following binary tree:

        3
       / 
      9  20
        /  
       15   7
    /**
     * Definition for a binary tree node.
     * public class TreeNode {
     *     int val;
     *     TreeNode left;
     *     TreeNode right;
     *     TreeNode(int x) { val = x; }
     * }
     */
    class Solution {
        public TreeNode buildTree(int[] preorder, int[] inorder) {
            Map<Integer, Integer> mymap = new HashMap<>();
            for (int i = 0; i < inorder.length; i++) {
                mymap.put(inorder[i], i);
            }
            return helper(0, inorder.length - 1, 0, preorder.length - 1, preorder, mymap);
        }
        
        private TreeNode helper(int inLeft, int inRight, int preLeft, int preRight, int[] preorder, Map<Integer, Integer> mymap) {
            if (inLeft > inRight) {
                return null;
            }
            TreeNode cur = new TreeNode(preorder[preLeft]);
            int leftSize = mymap.get(preorder[preLeft]) - inLeft; 
            // preRight for left just add up leftSize
            cur.left = helper(inLeft, inLeft + leftSize - 1, preLeft + 1, preLeft + leftSize, preorder, mymap);
            cur.right = helper(inLeft + leftSize + 1, inRight, preLeft + leftSize + 1, preRight, preorder, mymap);
            return cur;
        }
    }
     
     
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  • 原文地址:https://www.cnblogs.com/xuanlu/p/12170809.html
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