September 23, 2019

105. Construct Binary Tree from Preorder and Inorder Traversal

105. Construct Binary Tree from Preorder and Inorder Traversal

Similar question:

Time = O(n), Space = O(n)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */
class Solution {
    public TreeNode buildTree(int[] preorder, int[] inorder) {
        Map<Integer, Integer> inMap = new HashMap<>();
        for (int i = 0; i < inorder.length; i++) {
            inMap.put(inorder[i], i);
        }
        return buildTree(preorder, 0, preorder.length - 1, inorder, 0, inorder.length - 1, inMap);
    }
    
    public TreeNode buildTree(int[] preorder, int preStart, int preEnd, int[] inorder, int inStart, int inEnd, Map<Integer, Integer> inMap) {
        if (preStart > preEnd || inStart > inEnd) {
            return null;
        }
        TreeNode root = new TreeNode(preorder[preStart]);
        int inRoot = inMap.get(root.val);
        int numsLeft = inRoot - inStart;
        
        root.left = buildTree(preorder, preStart + 1, preStart + numsLeft, inorder, inStart, inRoot - 1, inMap);
        root.right = buildTree(preorder, preStart + numsLeft + 1, preEnd, inorder, inRoot + 1, inEnd, inMap);
        return root;
    }
}
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