Leetcode 从中序与后序遍历序列构造二叉树

java 递归实现

class Solution {
    private HashMap<Integer, Integer> inorderIndexMap;
    private int postorderIndex;

    public TreeNode buildTree(int[] inorder, int[] postorder) {
        inorderIndexMap = new HashMap<>();
        postorderIndex = postorder.length - 1;

        for(int i = 0; i < inorder.length; i++) {
            inorderIndexMap.put(inorder[i], i);
        }

        return Helper(postorder, 0, inorder.length - 1);
    }

    private TreeNode Helper(int[] postorder, int inorderStart, int inorderEnd) {
        if(inorderStart > inorderEnd) return null;
        //首先确定根节点的值
        int rootValue = postorder[postorderIndex--];
        TreeNode root = new TreeNode(rootValue);

        //获取根节点在中序遍历中的索引，用于后面划分左右子树
        int inorderIndex = inorderIndexMap.get(rootValue);

        // 递归构建右子树和左子树
        // 注意：需要先构建右子树，因为后序遍历的顺序是左右根
        root.right = Helper(postorder, inorderIndex + 1, inorderEnd);
        root.left = Helper(postorder, inorderStart, inorderIndex - 1);

        return root;
    }
}