LC105&106: construct BT from Preorder & Inorder/Inorder & Postorder

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode(int x) { val = x; }
 * }
 */

细细体会，其实关键点也不难，就是根据inorder里面的确定的left/right subtree对应的node的个数分别是多少，然后再在preorder/postorder中找到和根节点不相邻的部分的range。

105. Construct Binary Tree from Preorder and Inorder Traversal

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

106. Construct Binary Tree from Inorder and Postorder Traversal

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