LeetCode Construct Binary Tree from Preorder and Inorder Traversal

最新推荐文章于 2024-01-17 14:50:43 发布

原创最新推荐文章于 2024-01-17 14:50:43 发布 · 2.1k 阅读

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本文介绍了一种使用前序遍历和中序遍历构建二叉树的方法。通过递归算法和哈希映射来确定每个节点的位置，最终完成整棵树的构建。

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

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

思路分析：这题类似于LeetCode Construct Binary Tree from Inorder and Postorder Traversal，可以参考这篇题解，主要考察递归。

AC Code

/**
 * Definition for binary tree
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        HashMap<Integer, Integer> inorderValueToIndexMap = new HashMap<Integer, Integer>();
        for(int i = 0; i < inorder.length; i++){
            inorderValueToIndexMap.put(inorder[i], i);
        }
        return buildTreeHelper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, inorderValueToIndexMap);
    }
    
    public TreeNode buildTreeHelper(int [] preorder, int ps, int pe, int[] inorder, int is, int ie, HashMap<Integer, Integer> inorderValueToIndexMap){
        if(ps > pe || is > ie) return null;
        TreeNode root = new TreeNode(preorder[ps]);
        int indexOfRootInorder = inorderValueToIndexMap.get(root.val);
        int leftNum = indexOfRootInorder - is;
        int rightNum = ie - indexOfRootInorder;
        root.left = buildTreeHelper(preorder, ps + 1, ps + leftNum, inorder, indexOfRootInorder - leftNum, indexOfRootInorder -1, inorderValueToIndexMap);
        root.right = buildTreeHelper(preorder, ps + leftNum + 1, ps + leftNum + rightNum, inorder, indexOfRootInorder + 1, indexOfRootInorder + rightNum, inorderValueToIndexMap);
        return root;
    }
}