Construct Binary Tree from Inorder and Postorder Traversal
最后更新于：2022-04-02 01:11:34
# Construct Binary Tree from Inorder and Postorder Traversal
### Source
- lintcode: [(72) Construct Binary Tree from Inorder and Postorder Traversal](http://www.lintcode.com/en/problem/construct-binary-tree-from-inorder-and-postorder-traversal/)
~~~
Given inorder and postorder traversal of a tree, construct the binary tree.
Example
Given inorder [1,2,3] and postorder [1,3,2], return a tree:
2
/ \
1 3
Note
You may assume that duplicates do not exist in the tree.
~~~
### 题解
和题 [Construct Binary Tree from Preorder and Inorder Traversal](http://algorithm.yuanbin.me/zh-cn/binary_tree/construct_binary_tree_from_preorder_and_inorder_traversal.html) 几乎一致，关键在于找到中序遍历中的根节点和左右子树，递归解决。
### Java
~~~
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
*@param inorder : A list of integers that inorder traversal of a tree
*@param postorder : A list of integers that postorder traversal of a tree
*@return : Root of a tree
*/
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (inorder == null || postorder == null) return null;
if (inorder.length == 0 || postorder.length == 0) return null;
if (inorder.length != postorder.length) return null;
TreeNode root = helper(inorder, 0, inorder.length - 1,
postorder, 0, postorder.length - 1);
return root;
}
private TreeNode helper(int[] inorder, int instart, int inend,
int[] postorder, int poststart, int postend) {
// corner cases
if (instart > inend || poststart > postend) return null;
// build root TreeNode
int root_val = postorder[postend];
TreeNode root = new TreeNode(root_val);
// find index of root_val in inorder[]
int index = findIndex(inorder, instart, inend, root_val);
// build left subtree
root.left = helper(inorder, instart, index - 1,
postorder, poststart, poststart + index - instart - 1);
// build right subtree
root.right = helper(inorder, index + 1, inend,
postorder, poststart + index - instart, postend - 1);
return root;
}
private int findIndex(int[] nums, int start, int end, int target) {
for (int i = start; i <= end; i++) {
if (nums[i] == target) return i;
}
return -1;
}
}
~~~
### 源码分析
找根节点的方法作为私有方法，辅助函数需要注意索引范围。
### 复杂度分析
找根节点近似 O(n)O(n)O(n), 递归遍历整个数组，嵌套找根节点的方法，故总的时间复杂度为 O(n2)O(n^2)O(n2).
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