106. Construct Binary Tree from

题目链接

tag:

• Medium；

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

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

Example:

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]
Return the following binary tree:
3
/ \
9 20
 / \
15 7

思路：
  思路同105. Construct Binary Tree from Preorder and Inorder Traversal一摸一样，后序遍历的最后一个肯定是根，所以原二叉树的根节点可以知道，题目中给了一个很关键的条件就是树中没有相同元素，有了这个条件我们就可以在中序遍历中也定位出根节点的位置，并以根节点的位置将中序遍历拆分为左右两个部分，分别对其递归调用原函数即可。代码如下：

/**
 * Definition for a binary tree node.
 * struct TreeNode {
 *     int val;
 *     TreeNode *left;
 *     TreeNode *right;
 *     TreeNode(int x) : val(x), left(NULL), right(NULL) {}
 * };
 */
class Solution {
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return buildTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() -1);
    }
    
    TreeNode* buildTree(vector<int>& inorder, int inLeft, int inRight, vector<int>& postorder, int postLeft, int postRight) {
        if (inLeft > inRight || postLeft > postRight) return NULL;
        int i = 0;
        for (i=inLeft; i<=inRight; ++i) {
            if (inorder[i] == postorder[postRight]) break;
        }
        
        TreeNode *cur = new TreeNode(postorder[postRight]);
        cur->left = buildTree(inorder, inLeft, i - 1, postorder, postLeft, postLeft + i - inLeft - 1);
        cur->right = buildTree(inorder, i + 1, inRight, postorder, postLeft + i - inLeft, postRight - 1);
        return cur;
    }
};