力扣面试150 从中序与后序遍历序列构造二叉树 递归

Problem: 106. 从中序与后序遍历序列构造二叉树

👨‍🏫 参考题解

🍻 Code 1

/**
 * 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[] inorder, int[] postorder) {
        int n = postorder.length;
        if(n == 0){
            return null;
        }
        int leftSize = indexOf(inorder, postorder[n-1]);// 左区间的长度
        int[] in1 = Arrays.copyOfRange(inorder, 0, leftSize);
        int[] in2 = Arrays.copyOfRange(inorder, leftSize+1, n);// 记得排除中间的根节点
        int[] post1 = Arrays.copyOfRange(postorder, 0, leftSize);
        int[] post2 = Arrays.copyOfRange(postorder, leftSize, n-1);// 排除最后一个根节点
        TreeNode left = buildTree(in1, post1);
        TreeNode right = buildTree(in2,post2);
        return new TreeNode(postorder[n-1],left,right);
    }

    // 返回 x 在 a 中的下标，保证 x 一定在 a 中
    private int indexOf(int[] a , int x){
        for(int i = 0; ; i++){
            if(a[i] == x){
                return i;
            }
        }
    }
}

🍻 哈希优化

class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        int n = inorder.length;
        Map<Integer, Integer> index = new HashMap<>(n); // 预分配空间
        for (int i = 0; i < n; i++) {
            index.put(inorder[i], i);
        }
        return dfs(inorder, 0, n, postorder, 0, n, index); // 左闭右开区间
    }

    private TreeNode dfs(int[] inorder, int inL, int inR, int[] postorder, int postL, int postR, Map<Integer, Integer> index) {
        if (postL == postR) { // 空节点
            return null;
        }
        int leftSize = index.get(postorder[postR - 1]) - inL; // 左子树的大小
        TreeNode left = dfs(inorder, inL, inL + leftSize, postorder, postL, postL + leftSize, index);
        TreeNode right = dfs(inorder, inL + leftSize + 1, inR, postorder, postL + leftSize, postR - 1, index);
        return new TreeNode(postorder[postR - 1], left, right);
    }
}