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());
}
};