二叉树重建

LeetCode 105.

给定两个整数数组preOrder 和inOrder，其中preOrder是二叉树的先序遍历，inOrder是二叉树的中序遍历，请构造二叉树并返回其根节点

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    Dictionary<int, int> inOrderMap = new Dictionary<int, int>();

    public TreeNode BuildTree(int[] preOrder, int[] inOrder) {
        for (int i = 0; i < inOrder.Length; i++)
            inOrderMap.Add(inOrder[i], i);

        return BuildTreeByRecursion(preOrder, 0, preOrder.Length - 1, inOrder, 0, inOrder.Length - 1);
    }

    public TreeNode BuildTreeByRecursion(int[] preOrder, int preStart, int preEnd, int[] inOrder, int inStart, int inEnd)
    {
        //先序遍历的第一个节点一定是当前子树的根节点
        int rootValue = preOrder[preStart];
        int rootInOrderIndex = inOrderMap[rootValue];

        TreeNode root = new TreeNode(rootValue);
　　　　 //当前子树左孩子节点数
        int leftNodes = rootInOrderIndex - inStart;
        
　　　　  //当前子树右孩子节点数
　　　　  int rightNodes = inEnd - rootInOrderIndex;

        if (leftNodes > 0)
            root.left = BuildTreeByRecursion(preOrder, preStart + 1, preStart + leftNodes, inOrder, inStart, rootInOrderIndex - 1);
        
        if (rightNodes > 0)
            root.right = BuildTreeByRecursion(preOrder, preStart + leftNodes + 1, preEnd, inOrder, rootInOrderIndex + 1, inEnd);

        return root;
    }
}

 LeetCode 106.

给定两个整数数组inOrder 和postOrder，其中inOrder是二叉树的中序遍历，postOrder是二叉树的后序遍历，请构造二叉树并返回其根节点

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     public int val;
 *     public TreeNode left;
 *     public TreeNode right;
 *     public TreeNode(int val=0, TreeNode left=null, TreeNode right=null) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
public class Solution {
    Dictionary<int, int> inOrderMap = new Dictionary<int, int>();

    public TreeNode BuildTree(int[] inOrder, int[] postOrder) {
        for (int i = 0; i < inOrder.Length; i++)
            inOrderMap.Add(inOrder[i], i);

        return BuildTreeByRecursion(inOrder, 0, inOrder.Length - 1, postOrder, 0, postOrder.Length - 1);
    }

    public TreeNode BuildTreeByRecursion(int[] inOrder, int inStart, int inEnd, int[] postOrder, int postStart, int postEnd)
    {
　　　　  //后续遍历的最后一个节点一定是当前子树的根节点
        int rootValue = postOrder[postEnd];
        int rootInOrderIndex = inOrderMap[rootValue];

        TreeNode root = new TreeNode(rootValue);
        int leftNodes = rootInOrderIndex - inStart;
        int rightNodes = inEnd - rootInOrderIndex;

        if (leftNodes > 0)
            root.left = BuildTreeByRecursion(inOrder, inStart, rootInOrderIndex - 1, postOrder, postStart, postStart + leftNodes - 1);
        
        if (rightNodes > 0)
            root.right = BuildTreeByRecursion(inOrder, rootInOrderIndex + 1, inEnd, postOrder, postEnd - rightNodes, postEnd - 1);

        return root;
    }
}