105. 从前序与中序遍历序列构造二叉树

递归

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        /**
        先序：中左右 中序： 左中右 后续：左右中
        先序遍历第一位为根节点
        // 根据 idx 来递归找左右子树
        root.left = helper(preorder, preLeft + 1, preLeft + (idx - inLeft),
                         inorder, inLeft, idx - 1);
        root.right = helper(preorder, preLeft + (idx - inLeft) + 1, preRight,
                         inorder, idx + 1, inRight);
         */
        return buildTree(preorder,0,preorder.length,
        inorder,0,inorder.length);
    }
    public TreeNode buildTree(int[] preorder,int preL,int preR,
    int[] inorder, int inL, int inR ) {
        if(inR - inL < 1) return null;
        if(inR - inL == 1 ) return new TreeNode(inorder[inL]);
        int rootVal = preorder[preL];
        //根节点在中序数组中的位置
        int rootIndex = inL;
        for(int i = inL; i < inR; i++ ) {
            if(inorder[i] == rootVal ) {
                rootIndex = i;
                break;
            }
        }
        TreeNode root = new TreeNode(rootVal);
        root.left = buildTree(preorder,preL + 1,preL + (rootIndex - inL),
        inorder,inL,rootIndex);
        root.right = buildTree(preorder,preL + (rootIndex - inL) + 1, preR,
        inorder,rootIndex + 1,inR);
        return root;
    }
}