从前序与中序遍历序列构造二叉树-二叉树

leetcode-105 从前序与中序遍历序列构造二叉树

前序的第一个节点是根节点，然后在中序中找到这个节点的位置，该位置前为左子树，后为右子树。

/**
 * 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 preLeft, int preRight, int[] inorder, int inLeft, int inRight){
        if (preLeft > preRight) return null;
        int i = 0;
        TreeNode root = new TreeNode(preorder[preLeft]);
        // 查找根节点在中序中的位置
        for (i = inLeft; i <= inRight; i++){
            if (inorder[i] == preorder[preLeft]){
                break;
            }
        }
        // 构建左子树
        root.left = buildTree(preorder, preLeft + 1, preLeft + (i - inLeft), inorder, inLeft, i - 1);
        // 构建右子树
        root.right = buildTree(preorder, preLeft + 1 + (i - inLeft), preRight, inorder, i + 1, inRight);
        return root;
    }
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        TreeNode root = buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1);
        return root;
    }
}

本文地址：https://blog.csdn.net/liu_tian_wei/article/details/109637936