Given inorder and postorder traversal of a tree, construct the binary tree.
Notice
You may assume that duplicates do not exist in the tree.
Given inorder [1,2,3]
and
postorder [1,3,2]
, return a tree:
2
/ \
1 3
/**
* Definition of TreeNode:
* public class TreeNode {
* public int val;
* public TreeNode left, right;
* public TreeNode(int val) {
* this.val = val;
* this.left = this.right = null;
* }
* }
*/
public class Solution {
/**
*@param inorder : A list of integers that inorder traversal of a tree
*@param postorder : A list of integers that postorder traversal of a tree
*@return : Root of a tree
*/
public TreeNode buildTree(int[] inorder, int[] postorder) {
if(inorder.length == 0 || postorder.length == 0) {
return null;
}
return construct(inorder, 0, inorder.length - 1, postorder, 0, postorder.length - 1);
}
private TreeNode construct(int[] inorder, int inStart, int inEnd, int[] postorder, int postStart, int postEnd) {
if(inStart > inEnd || postStart > postEnd) {
return null;
}
TreeNode root = new TreeNode(postorder[postEnd]);
int position = 0;
for(int i = 0; i < inorder.length; i++) {
if(inorder[i] == postorder[postEnd]) {
position = i;
}
}
int leftLength = position - inStart;
root.left = construct(inorder, inStart, position - 1, postorder, postStart, postStart + leftLength - 1);//be careful about the index
root.right = construct(inorder, position + 1, inEnd, postorder, postStart + leftLength, postEnd - 1);
return root;
}
}