Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
题意：给出一个二叉树的中序和后序遍历结果，并且这个二叉树中没有重复值，复原这个二叉树。
思路：这道题和105题相比，只是条件从前序换成了后序，但是后序数组的最后一个值仍然是根节点，因此依旧可以用105题的解法解决。
public TreeNode buildTree(int[] inorder, int[] postorder) {
if (inorder == null || postorder == null || inorder.length != postorder.length) {
return null;
}
return helper(0, inorder.length - 1, 0, postorder.length - 1, inorder, postorder);
}
public TreeNode helper(int inStart, int inEnd, int postStart, int postEnd, int[] inorder, int[] postorder) {
if (inStart > inEnd || postStart > postEnd) {
return null;
}
//find parent
TreeNode parent = new TreeNode(postorder[postEnd]);
//find left right in inorder
int inIndex = inStart;
for (int i = inStart; i <= inEnd; i++) {
if (inorder[i] == postorder[postEnd]) {
inIndex = i;
break;
}
}
parent.left = helper(inStart, inIndex - 1, postStart, postStart + inIndex - inStart - 1, inorder, postorder);
parent.right = helper(inIndex + 1, inEnd, postStart + inIndex - inStart, postEnd - 1, inorder, postorder);
return parent;
}