leetcode105. Construct Binary Tree from Preorder and Inorder Traversal

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本文介绍了一种从给定的前序和中序遍历序列构建二叉树的方法。通过递归地寻找根节点，并确定左右子树的范围，最终构建完整的二叉树结构。

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.

解法一

前序遍历的第一个节点就是根节点，在中序遍历中找到根节点所在的位置，比如M，中序遍历的0 - M-1的节点属于左子树，M+1 - inorder.length - 1属于右子树。
按照该方法进行dfs。

/**
 * 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 || preorder.length == 0) && (inorder == null || inorder.length == 0)){
            return null;
        }

        return helper(0, 0, inorder.length - 1, preorder, inorder);
    }
    private TreeNode helper(int preStart, int inStart, int inEnd, int[] preorder, int[] inorder) {
        if (preStart > preorder.length - 1 || inStart > inEnd) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[preStart]);
        int inIndex = 0; // index of current root in inorder
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == root.val) {
                inIndex = i;
            }
        }
        root.left = helper(preStart + 1, inStart, inIndex - 1, preorder, inorder);
        root.right = helper(preStart + inIndex - inStart + 1, inIndex + 1, inEnd, preorder, inorder);

        return root;
    }
}

解法二

/**
 * 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 || preorder.length == 0) && (inorder == null || inorder.length == 0)){
            return null;
        }

        return helper(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
    }
    private TreeNode helper(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd) {
        if (preStart > preEnd || inStart > inEnd) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[preStart]);
        int inIndex = 0; // index of current root in inorder
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == root.val) {
                inIndex = i;
            }
        }
        root.left = helper(preorder, preStart + 1, preStart + inIndex - inStart, inorder, inStart, inIndex - 1);
        root.right = helper(preorder, preStart + inIndex - inStart + 1, preEnd, inorder, inIndex + 1, inEnd);

        return root;
    }
}