Leetcode: Construct Binary Tree from Inorder and Postorder Traversal

题目：

Given inorder and postorder traversal of a tree, construct the binary tree.

Note:

You may assume that duplicates do not exist in the tree.

根据二叉树的中序遍历和后续遍历结果构造二叉树。

思路分析：

这道题和上道题《 Leetcode: Construct Binary Tree from Preorder and Inorder Traversal 》的思路类似。

二叉树后序遍历的最后一个节点是根节点。

在中序遍历中找出根节点，由根节点分开，中序遍历左边是左子树中序遍历结果（设有X个节点），右边是右子树遍历结果（设有Y个节点）。

后序遍历除去最后一个根节点，数X节点，这X个节点是左子树后序遍历的结果，剩下节点（肯定是Y个）是右子树后序遍历结果。

这样我们就得到了左右子树的后序遍历和中序遍历的结果了，所以又回到了原来的问题。

C++参考代码：

/**
* Definition for binary tree
* struct TreeNode {
*     int val;
*     TreeNode *left;
*     TreeNode *right;
*     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
TreeNode *makeNode(vector<int>::iterator inBegin, vector<int>::iterator inEnd, vector<int>::iterator postBegin, vector<int>::iterator postEnd)
{
if (postBegin == postEnd) return nullptr;
//从中序遍历结果中找出根节点（根节点在后序遍历中是最后一个节点）
vector<int>::iterator itRoot = find(inBegin, inEnd, *(postEnd - 1));
TreeNode *root = new TreeNode(*itRoot);
//计算根的左子树节点个数
int leftSize = itRoot - inBegin;
//在中序遍历结果中根节点的左边是左子树中序遍历结果，右边是右子树中序遍历结果
//在后序遍历结果中除去根节点前leftSize个节点是左子树前序遍历结果，后面的节点是右子树前序遍历结果
root->left = makeNode(inBegin, itRoot, postBegin, postBegin + leftSize);
root->right = makeNode(itRoot + 1, inEnd, postBegin + leftSize, postEnd - 1);
return root;
}

public:
TreeNode *buildTree(vector<int> &inorder, vector<int> &postorder)
{
if (postorder.empty()) return nullptr;
TreeNode *root = makeNode(inorder.begin(), inorder.end(), postorder.begin(), postorder.end());
return root;
}
};

Java参考代码：

/**
* Definition for binary tree
* public class TreeNode {
*     int val;
*     TreeNode left;
*     TreeNode right;
*     TreeNode(int x) { val = x; }
* }
*/
public class Solution {
private TreeNode makeNode(int[] inorder, int inBegin, int inEnd, int[] postorder, int postBegin, int postEnd) {
if (postBegin == postEnd) return null;
int index = 0;
for (int i = inBegin; i < inEnd; i++) {
if (inorder[i] == postorder[postEnd - 1]) {
index = i;
}
}
int leftSize = index - inBegin;
TreeNode root = new TreeNode(postorder[postEnd - 1]);
root.left = makeNode(inorder, inBegin, inBegin + leftSize, postorder, postBegin, postBegin + leftSize);
root.right = makeNode(inorder, index + 1, inEnd, postorder, postBegin + leftSize, postEnd - 1);
return root;
}

public TreeNode buildTree(int[] inorder, int[] postorder) {
if (postorder.length == 0) return null;
TreeNode root = makeNode(inorder, 0, inorder.length, postorder, 0, postorder.length);
return root;
}
}