Construct Binary Tree from Inorder and Postorder Traversal
Problem Description​
Given two integer arrays inorder and postorder where inorder is the inorder traversal of a binary tree and postorder is the postorder traversal of the same tree, construct and return the binary tree.
Examples​
Example 1​
- Input:
inorder = [9,3,15,20,7], postorder = [9,15,7,20,3] - Output:
[3,9,20,null,null,15,7]
Example 2​
- Input:
inorder = [-1], postorder=[-1] - Output:
[-1]
Constraints​
- postorder and inorder consist of unique values.
- Each value of postorder also appears in inorder.
- postorder is guaranteed to be the postorder traversal of the tree.
- inorder is guaranteed to be the inorder traversal of the tree.
Solution Code​
Python​
class Solution:
def buildTree(self, inorder: List[int], postorder: List[int]) -> Optional[TreeNode]:
n=len(inorder)
poststart=instart=0
postend=inend=n-1
d={}
for i in range(n):
d[inorder[i]]=i
return self.constructTree(postorder,poststart,postend,inorder,instart,inend,d)
def constructTree(self,postorder,poststart,postend,inorder,instart,inend,d):
if poststart>postend or instart>inend:
return None
root=TreeNode(postorder[postend])
elem=d[root.val]
nelem=elem-instart
root.left=self.constructTree(postorder,poststart,poststart+nelem-1,inorder,instart,elem-1,d)
root.right=self.constructTree(postorder,poststart+nelem,postend-1,inorder,elem+1,inend,d)
return root
Java​
class Solution
{
public TreeNode buildTree(int[] inorder, int[] postorder) {
// If either of the input arrays are empty, the tree is empty, so return null
if (inorder.length == 0 || postorder.length == 0) return null;
// Initialize indices to the last elements of the inorder and postorder traversals
int ip = inorder.length - 1;
int pp = postorder.length - 1;
// Create an empty stack to help us build the binary tree
Stack<TreeNode> stack = new Stack<TreeNode>();
// Initialize prev to null since we haven't processed any nodes yet
TreeNode prev = null;
// Create the root node using the last element in the postorder traversal
TreeNode root = new TreeNode(postorder[pp]);
// Push the root onto the stack and move to the next element in the postorder traversal
stack.push(root);
pp--;
// Process the rest of the nodes in the postorder traversal
while (pp >= 0) {
// While the stack is not empty and the top of the stack is the current inorder element
while (!stack.isEmpty() && stack.peek().val == inorder[ip]) {
// The top of the stack is the parent of the current node, so pop it off the stack and update prev
prev = stack.pop();
ip--;
}
// Create a new node for the current postorder element
TreeNode newNode = new TreeNode(postorder[pp]);
// If prev is not null, the parent of the current node is prev, so attach the node as the left child of prev
if (prev != null) {
prev.left = newNode;
// If prev is null, the parent of the current node is the current top of the stack, so attach the node as the right child of the current top of the stack
} else if (!stack.isEmpty()) {
TreeNode currTop = stack.peek();
currTop.right = newNode;
}
// Push the new node onto the stack, reset prev to null, and move to the next element in the postorder traversal
stack.push(newNode);
prev = null;
pp--;
}
// Return the root of the binary tree
return root;
}
}
C++​
class Solution {
public:
TreeNode* buildTree(vector<int>& inorder, vector<int>& postorder) {
// If either of the input vectors are empty, the tree is empty, so return null
if (inorder.size() == 0 || postorder.size() == 0) return nullptr;
// Initialize indices to the last elements of the inorder and postorder traversals
int ip = inorder.size() - 1;
int pp = postorder.size() - 1;
// Create an empty stack to help us build the binary tree
stack<TreeNode*> st;
// Initialize prev to null since we haven't processed any nodes yet
TreeNode* prev = nullptr;
// Create the root node using the last element in the postorder traversal
TreeNode* root = new TreeNode(postorder[pp]);
// Push the root onto the stack and move to the next element in the postorder traversal
st.push(root);
pp--;
// Process the rest of the nodes in the postorder traversal
while (pp >= 0) {
// While the stack is not empty and the top of the stack is the current inorder element
while (!st.empty() && st.top()->val == inorder[ip]) {
// The top of the stack is the parent of the current node, so pop it off the stack and update prev
prev = st.top();
st.pop();
ip--;
}
// Create a new node for the current postorder element
TreeNode* newNode = new TreeNode(postorder[pp]);
// If prev is not null, the parent of the current node is prev, so attach the node as the left child of prev
if (prev != nullptr) {
prev->left = newNode;
// If prev is null, the parent of the current node is the current top of the stack, so attach the node as the right child of the current top of the stack
} else if (!st.empty()) {
TreeNode* currTop = st.top();
currTop->right = newNode;
}
// Push the new node onto the stack, reset prev to null, and move to the next element in the postorder traversal
st.push(newNode);
prev = nullptr;
pp--;
}
// Return the root of the binary tree
return root;
}
};
Javascript​
var buildTree = function(inorder, postorder) {
if (inorder.length == 0 || postorder.length == 0) {
return null;
}
var rootVal = postorder[postorder.length - 1];
var root = new TreeNode(rootVal);
var rootIndex = inorder.indexOf(rootVal);
var leftInorder = inorder.slice(0, rootIndex);
var rightInorder = inorder.slice(rootIndex + 1);
var leftPostorder = postorder.slice(0, leftInorder.length);
var rightPostorder = postorder.slice(leftInorder.length, postorder.length - 1);
root.left = buildTree(leftInorder, leftPostorder);
root.right = buildTree(rightInorder, rightPostorder);
return root;
};