LeetCode 106: Construct Binary Tree from Inorder and Postorder Traversal

Construct Binary Tree from Inorder and Postorder Traversal

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

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

解题思路

方法和 LeetCode 105 类似，不过这里要利用后序序列的最后一个数字一定是树的根的性质。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 {
    TreeNode* buildTree(vector<int>& inorder, int il, int ir, 
                        vector<int>& postorder, int pl, int pr) {
        if (pl >= pr) return NULL;

        // 构造根节点
        TreeNode *root = new TreeNode(postorder[pr-1]);

        // 左子树大小
        int leftSize = 0;
        while (inorder[il + leftSize] != postorder[pr-1]) leftSize++;

        // 递归构造左子树
        root->left = buildTree(inorder, il, il + leftSize, postorder, pl, pl + leftSize);
        // 递归构造右子树
        root->right = buildTree(inorder, il + leftSize + 1, ir, 
                                    postorder, pl + leftSize, pr - 1);

        return root;
    }
public:
    TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
        return buildTree(inorder, 0, inorder.size(), postorder, 0, postorder.size());
    }
};