Construct Binary Tree from Inorder and Preorder Traversal

Problem

Given preorder and inorder traversal of a tree, construct the binary tree.

Notice

You may assume that duplicates do not exist in the tree.

Example

Given in-order [1,2,3] and pre-order [2,1,3], return a tree:

  2
 / \
1   3

Note

许久未更。做了几道二分法的题目练手，发现这道题已经淡忘了，记录一下。

这道题目的要点在于找inorder的区间。preStart每增加一次，就对应一个新的子树。每个子树的根节点都可以在inorder的中间某处找到，以此为界，左边是这个根节点的左子树，右边是右子树。不断递归，得解。

边界条件需要注意：

1. 若preorder或inorder数组为空，返回空；

2. 当preStart前进到超出preorder末位，或inStart超过inEnd，返回空；

3. 每次创建完根节点之后，要将preStart加1，才能进行递归。

Solution

public class Solution {
    int preStart = 0;
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        if (preorder.length == 0 || inorder.length == 0) return null;
        return helper(preorder, inorder, 0, preorder.length-1);
    }
    public TreeNode helper(int[] preorder, int[] inorder, int inStart, int inEnd) {
        if (preStart >= preorder.length || inStart > inEnd) return null;
        int index = 0;
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == preorder[preStart]) {
                index = i;
                break;
            }
        }
        TreeNode root = new TreeNode(preorder[preStart++]);
        root.left = helper(preorder, inorder, inStart, index-1);
        root.right = helper(preorder, inorder, index+1, inEnd);
        return root;
    }
}

Construct Binary Tree from Inorder and Postorder Traversal

Problem

Given inorder and postorder traversal of a tree, construct the binary tree.

Notice

You may assume that duplicates do not exist in the tree.

Example

Given inorder [1,2,3] and postorder [1,3,2], return a tree:

  2
 / \
1   3

Note

和先序+中序方法一样，不过这次是逆推，递归的时候先右子树，后左子树即可。

Solution

Recursion

public class Solution {
    int postEnd;
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        postEnd = postorder.length - 1;
        return helper(postorder, inorder, 0, inorder.length - 1);
    }
    
    private TreeNode helper(int[] postorder, int[] inorder, int inStart, int inEnd) {
        if (postEnd < 0 || inStart > inEnd) return null;
        TreeNode root = new TreeNode(postorder[postEnd--]);
        
        int inMid = 0;
        for (int i = inStart; i <= inEnd; i++) {
            if (inorder[i] == root.val) {
                inMid = i;
                break;
            }
        }
        root.right = helper(postorder, inorder, inMid +1, inEnd);
        root.left = helper(postorder, inorder, inStart, inMid-1);
        return root;
    }
}

Using Stack

public class Solution {
    public TreeNode buildTree(int[] inorder, int[] postorder) {
        if (inorder == null || inorder.length < 1) return null;
        int i = inorder.length - 1;
        int p = i;
        TreeNode node;
        TreeNode root = new TreeNode(postorder[postorder.length - 1]);
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root);
        p--;
        while (true) {
            if (inorder[i] == stack.peek().val) { // inorder[i] is on top of stack, pop stack to get its parent to get to left side
                if (--i < 0) break;
                node = stack.pop();
                if (!stack.isEmpty() && inorder[i] == stack.peek().val) continue;
                node.left = new TreeNode(postorder[p]);
                stack.push(node.left);
            } else { // inorder[i] is not on top of stack, postorder[p] must be right child
                node = new TreeNode(postorder[p]);
                stack.peek().right = node;
                stack.push(node);
            }
            p--;
        }
        return root;
    }
}