【二叉树】106. 从中序与后序遍历序列构造二叉树

题目

法1：递归构造

Python

# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, val=0, left=None, right=None):
#         self.val = val
#         self.left = left
#         self.right = right
class Solution:
    def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
        if len(inorder) == 0:
            return None
        elif len(inorder) == 1:
            return TreeNode(inorder[0])
        root_val = postorder[-1]
        left_size = inorder.index(root_val)
        left = self.buildTree(inorder[: left_size], postorder[: left_size])
        right = self.buildTree(inorder[left_size+1:], postorder[left_size: -1])

        return TreeNode(root_val, left, right)

Java

class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        return build(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
    }

    public TreeNode build(int[] inorder, int inStart, int inEnd, int[] postorder, int postStart, int postEnd) {
        if (postStart > postEnd) {
            return null;
        }
        if (postStart == postEnd) {
            return new TreeNode(postorder[postEnd]);
        }

        TreeNode root = new TreeNode(postorder[postEnd]);
        int valInx = -1;
        for (int i = inStart; i <= inEnd; ++i) {
            if (inorder[i] == postorder[postEnd]) {
                valInx = i;
                break;
            }
        }
        int leftSize = valInx - inStart;
        root.left = build(inorder, inStart, valInx - 1, postorder, postStart, postStart + leftSize - 1);
        root.right = build(inorder, valInx + 1, inEnd, postorder, postStart + leftSize, postEnd - 1);
        return root;
    }
}