[Leetcode Week14]Construct Binary Tree from Inorder and Postorder Traversal
Construct Binary Tree from Inorder and Postorder Traversal 题解
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Description
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
Solution
class Solution {
public:
TreeNode* getTreeNode(vector<int>& inOrder, vector<int>& postOrder,
int inStart, int inEnd, int postStart, int postEnd) {
TreeNode* resultNode = new TreeNode(postOrder[postEnd]);
if (postStart == postEnd)
return resultNode;
int i;
int inNodeVal = postOrder[postEnd];
for (i = inStart; i <= inEnd; i++) {
if (inNodeVal == inOrder[i])
break;
}
if (i > inStart)
resultNode -> left =
getTreeNode(inOrder, postOrder, inStart, i - 1, postStart, postStart + i - 1 - inStart);
if (i < inEnd)
resultNode -> right =
getTreeNode(inOrder, postOrder, i + 1, inEnd, postStart + i - inStart, postEnd - 1);
return resultNode;
}
TreeNode* buildTree(vector<int>& inOrder, vector<int>& postOrder) {
int size = inOrder.size();
if (size == 0)
return NULL;
return getTreeNode(inOrder, postOrder, 0, size - 1, 0, size - 1);
}
};
解题描述
这道题是经典的树构建问题,通过树的中序遍历和后序遍历结果来重建树,基本的算法是通过每次从后序遍历数组末端取出元素,这个元素为当前树的根,然后再在中序遍历结果中找到这个根,根的两边分别就是左右子树。对左右子树继续递归进行相同的操作,直到数组为空即可。
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