Construct Binary Tree from Preorder and Inorder Traversal

最新推荐文章于 2024-01-17 14:50:43 发布

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最新推荐文章于 2024-01-17 14:50:43 发布

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本文介绍如何根据给定的先序和中序遍历序列构建二叉树的方法。通过解析先序遍历得到根节点，并在中序遍历中找到该根节点的位置来划分左右子树，递归地构建整棵二叉树。

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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.

Supposed that we have a tree

Preorder : 1,2,4,5,3,6,7

Inorder : 4,2,5,1,6,3,7

We know that the first number of preorder traversal is the root of the binary tree, so that we can get the index of the root in the inorder traversal. According to the inorder traversal, we can split the binary tree into left subtree and right subtree.

leftSubtree: 4,2,5

rightSubtree: 6,3,7

Based on the length of the leftSubtree and rightSubtree, we can get the root for right subtree and left subtree from the preorder traversal respectively.

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
public class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder == null || inorder == null || preorder.length == 0 || inorder.length == 0)
            return null;
        
        int treeLength = preorder.length;
        return buildTreeHelper(preorder, 0, treeLength - 1, inorder, 0, treeLength - 1);
    }
    
    private TreeNode buildTreeHelper(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd) {
        
        if (preStart > preEnd || inStart > inEnd)
            return null;
            
        int rootVal = preorder[preStart];
        int rootIndex = 0;
        
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == rootVal) {
                rootIndex = i;
                break;
            }
        }
        
        TreeNode root = new TreeNode(rootVal);
        int leftLength = rootIndex - inStart;
        int rightLength = inEnd - rootIndex;
        root.left = buildTreeHelper(preorder, preStart + 1, preStart + leftLength, inorder, inStart, rootIndex - 1);
        root.right = buildTreeHelper(preorder, preStart + leftLength + 1, preStart + leftLength + rightLength, inorder, rootIndex + 1, rootIndex + rightLength);
        
        return root;
    }
}