Leetcode 106. Construct Binary Tree from Inorder and Postorder Traversal

题目链接：https://leetcode.cn/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

1 方法思想

2 代码实现

/**
 * 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 {
    Map<Integer, Integer> postorderMap = new HashMap<>();

    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (postorder == null || inorder == null) return null;
        for (int i = 0; i < inorder.length; i++) {
            postorderMap.put(inorder[i], i);
        }

        TreeNode root = nodeBuild(postorder, 0, postorder.length, inorder, 0, inorder.length);
        return root;
    }

    public TreeNode nodeBuild(int[] postorder, int posStart, int lenPosorder, int[] inorder, int inStart, int lenInorder) {
        if (posStart >= lenPosorder || inStart >= lenInorder) {
            return null;
        }
        int rootIndex = postorderMap.get(postorder[lenPosorder - 1]);
        TreeNode root = new TreeNode(inorder[rootIndex]);

        root.left = nodeBuild(postorder, posStart, posStart + rootIndex - inStart, inorder, inStart, rootIndex);
        root.right = nodeBuild(postorder, posStart + rootIndex - inStart, lenPosorder - 1, inorder, rootIndex + 1, lenInorder);
        return root;
    }
}
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3 复杂度分析

时间复杂度：
空间复杂度：

4 涉及到知识点

5 总结