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
/**
* Definition for a binary tree node.
* public class TreeNode {
* public int val;
* public TreeNode left;
* public TreeNode right;
* public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
* this.val = val;
* this.left = left;
* this.right = right;
* }
* }
*/
public class Solution {
public TreeNode BuildTree(int[] inorder, int[] postorder) {
return Helper(postorder.Length-1, 0, inorder.Length-1, inorder, postorder);
}
private TreeNode Helper(int postIndex, int inStart, int inEnd, int[] inorder, int[] postorder){
if(postIndex<0 || inStart > inEnd){
return null;
}
var root = new TreeNode(postorder[postIndex]);
int i= postorder.Length -1;
for(;i>=0;i--){
if(inorder[i]==root.val){
break;
}
}
root.left = Helper(postIndex - 1 - inEnd + i, inStart, i - 1, inorder, postorder);
root.right = Helper(postIndex - 1 , i + 1, inEnd, inorder, postorder);
return root;
}
}Time Complexity: O(n)
Space Complexity: O(n)
