106. Construct Binary Tree from Inorder and Postorder Traversal

[Tcdian]

106. Construct Binary Tree from Inorder and Postorder Traversal

根据一棵树的中序遍历与后序遍历构造二叉树。

Note

你可以假设树中没有重复的元素。

Example

For example, given

inorder = [9,3,15,20,7]
postorder = [9,15,7,20,3]

Return the following binary tree

    3
   / \
  9  20
    /  \
   15   7

[Tcdian]

Solution

• JavaScript Solution

/**
 * Definition for a binary tree node.
 * function TreeNode(val, left, right) {
 *     this.val = (val===undefined ? 0 : val)
 *     this.left = (left===undefined ? null : left)
 *     this.right = (right===undefined ? null : right)
 * }
 */
/**
 * @param {number[]} inorder
 * @param {number[]} postorder
 * @return {TreeNode}
 */
var buildTree = function(inorder, postorder) {
    if (inorder.length === 0) {
        return null;
    }
    const val = postorder[postorder.length - 1];
    const separator = inorder.indexOf(val);
    const node = new TreeNode(val);
    node.left = buildTree(inorder.slice(0, separator), postorder.slice(0, separator));
    node.right = buildTree(inorder.slice(separator + 1), postorder.slice(separator, postorder.length - 1));
    return node;
};

• TypeScript Solution

/**
 * Definition for a binary tree node.
 * class TreeNode {
 *     val: number
 *     left: TreeNode | null
 *     right: TreeNode | null
 *     constructor(val?: number, left?: TreeNode | null, right?: TreeNode | null) {
 *         this.val = (val===undefined ? 0 : val)
 *         this.left = (left===undefined ? null : left)
 *         this.right = (right===undefined ? null : right)
 *     }
 * }
 */

function buildTree(inorder: number[], postorder: number[]): TreeNode | null {
    if (inorder.length === 0) {
        return null;
    }
    const val = postorder[postorder.length - 1];
    const separator = inorder.indexOf(val);
    const node = new TreeNode(val);
    node.left = buildTree(inorder.slice(0, separator), postorder.slice(0, separator));
    node.right = buildTree(inorder.slice(separator + 1), postorder.slice(separator, postorder.length - 1));
    return node;
};