Construct Binary Tree from Inorder and Postorder Traversal

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后序遍历的最后一个点肯定是树根，然后在中序遍历里找到这个树根，如此，便得到两个子树，依次递归下去即可。

/** * Definition for binary tree * 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 build(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() - 1);    }    TreeNode *build(vector<int> &inorder, int i_s, int i_e, vector<int> &postorder, int p_s, int p_e) {        if (i_s > i_e)            return NULL;        if (i_s == i_e) {            TreeNode *t = new TreeNode(inorder[i_s]);            return t;        }        int root = postorder[p_e];        int index_inorder = 0;        for (; index_inorder < inorder.size(); ++index_inorder)            if (inorder[index_inorder] == root)                break;        TreeNode *t = new TreeNode(root);        t->left = build(inorder, i_s, index_inorder - 1, postorder, p_s, p_s + index_inorder - 1 - i_s);        t->right = build(inorder, index_inorder + 1, i_e, postorder, p_s + index_inorder - i_s, p_e - 1);        return t;    }};

http://oj.leetcode.com/problems/construct-binary-tree-from-inorder-and-postorder-traversal/

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