【leetcode hot 100 105】从前序与中序遍历序列构造二叉树

错误解法一：preorder[0]为根节点，在inorder中找到preorder[0]的位置numInorder，其左边为左子树，右边为右子树。利用Arrays.copyOfRange()函数来取数组子集。

/**
 * 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) {
        if(preorder.length==0){
            return null;
        }
        // 找到preorder[0]在inoder中的位置
        int numInoger;
        for(numInoger=0;numInoger<inorder.length;numInoger++){
            if(preorder[0]==inorder[numInoger]){
                break;
            }
        }
        int[] newPreorder = Arrays.copyOfRange(preorder,1,preorder.length);
        int[] newInoderLeft = Arrays.copyOfRange(inorder,0,numInoger);
        int[] newInoderRight = Arrays.copyOfRange(inorder,numInoger+1,inorder.length);
        TreeNode left = buildTree(newPreorder, newInoderLeft);
        TreeNode right = buildTree(newPreorder, newInoderRight);

        return new TreeNode(preorder[0], left, right);
    }
}

错误原因：

int[] newInoderRight = Arrays.copyOfRange(inorder,numInoger+1,inorder.length)：Arrays.copyOfRange()无法处理numInoger+1==inorder.length的情况。

解法一：由解法二发现的问题改进上述错误解法 =》速度慢！！！！

/**
 * 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) {
        if(preorder.length==0){
            return null;
        }
        // 找到preorder[0]在inoder中的位置
        int numInoger;
        for(numInoger=0;numInoger<inorder.length;numInoger++){
            if(preorder[0]==inorder[numInoger]){
                break;
            }
        }
        // 得到左子树中的节点数目
        int num_left = numInoger;
        int[] newPreorderLeft = Arrays.copyOfRange(preorder,1,1+numInoger);
        int[] newPreorderRight = Arrays.copyOfRange(preorder,1+numInoger,preorder.length);
        int[] newInoderLeft = Arrays.copyOfRange(inorder,0,numInoger);
        int[] newInoderRight = Arrays.copyOfRange(inorder,numInoger+1,inorder.length);
        TreeNode left = buildTree(newPreorderLeft, newInoderLeft);
        TreeNode right = buildTree(newPreorderRight, newInoderRight);

        return new TreeNode(preorder[0], left, right);
    }
}

解法二：preorder[0]为根节点，在inorder中找到preorder[0]的位置numInorder，其左边为左子树，右边为右子树。利用双指针函数来取数组子集。

/**
 * 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) {
        int n = preorder.length;
        return myBuildTree(preorder, inorder, 0, n - 1, 0, n - 1);
    }

    public TreeNode myBuildTree(int[] preorder, int[] inorder, int preorder_left, int preorder_right, int inorder_left, int inorder_right){
        if(preorder_left>preorder_right){
            return null;
        }
        // 找到preorder[preorder_left]在inoder中的位置
        int numInoger;
        for(numInoger=inorder_left;numInoger<=inorder_right;numInoger++){
            if(preorder[preorder_left]==inorder[numInoger]){
                break;
            }
        }
        
        TreeNode root = new TreeNode(preorder[preorder_left]);
        // 得到左子树中的节点数目
        int num_left = numInoger - inorder_left;
        root.left = myBuildTree(preorder, inorder, preorder_left+1, preorder_left+num_left, inorder_left, numInoger-1);
        root.right = myBuildTree(preorder, inorder,preorder_left+num_left+1, preorder_right, numInoger+1, inorder_right);
        return root;
    }
}

注意：

• 不能使用TreeNode left = myBuildTree(preorder, inorder, preorder_left+1, preorder_right, inorder_left, numInoger-1)和TreeNode right = myBuildTree(preorder, inorder, preorder_left+1, preorder_right, numInoger+1, inorder_right)，递归的不是preorder余下的所有数，而是要区分余下的数哪些在左子树+哪些在右子数：利用左子树中的节点数目num_left = numInoger - inorder_left来确定
• 速度慢 =》改进：构造哈希映射，帮助我们快速定位根节点HashMap(inorder[i], i)