LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和後序遍歷建立二叉樹 C++
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
For example, given
inorder = [9,3,15,20,7] postorder = [9,15,7,20,3]
Return the following binary tree:
3 / 9 20 / 15 7
中序、後序遍歷得到二叉樹，可以知道每一次新數組的最後一個數為當時子樹的根節點，每次根據中序遍歷的根節點的左右兩邊確定左右子樹，再對應後序的左右子樹，不停遞歸得到根節點，可以建立二叉樹。每次由循環得到根節點在中序數組中坐標i
由中序遍歷知：每一次inorder的左子樹範圍[ileft，i-1],右子樹範圍[i+1,iright]
LeetCode 106. Construct Binary Tree from Inorder and Postorder Traversal 由中序和後序遍歷建立二叉樹 C++
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