二叉搜索樹的第 k 個(gè)結(jié)點(diǎn)

題目：給定一棵二叉搜索樹，請(qǐng)找出其中的第k大的結(jié)點(diǎn)。

解題思路

如果按照中序遍歷的順序遍歷一棵二叉搜索樹，遍歷序列的數(shù)值是遞增排序的。只需要用中序遍歷算法遍歷一棵二叉搜索樹，就很容易找出它的第k大結(jié)點(diǎn)。

結(jié)點(diǎn)定義

private static class BinaryTreeNode {
    private int val;
    private BinaryTreeNode left;
    private BinaryTreeNode right;
    public BinaryTreeNode() {
    }
    public BinaryTreeNode(int val) {
        this.val = val;
    }
    @Override
    public String toString() {
        return val + "";
    }
}

代碼實(shí)現(xiàn)

public class Test63 {
    private static class BinaryTreeNode {
        private int val;
        private BinaryTreeNode left;
        private BinaryTreeNode right;
        public BinaryTreeNode() {
        }
        public BinaryTreeNode(int val) {
            this.val = val;
        }
        @Override
        public String toString() {
            return val + "";
        }
    }
    public static BinaryTreeNode kthNode(BinaryTreeNode root, int k) {
        if (root == null || k < 1) {
            return null;
        }
        int[] tmp = {k};
        return kthNodeCore(root, tmp);
    }
    private static BinaryTreeNode kthNodeCore(BinaryTreeNode root, int[] k) {
        BinaryTreeNode result = null;
        // 先成左子樹中找
        if (root.left != null) {
          result =  kthNodeCore(root.left, k);
        }
        // 如果在左子樹中沒有找到
        if (result == null) {
            // 說明當(dāng)前的根結(jié)點(diǎn)是所要找的結(jié)點(diǎn)
            if(k[0] == 1) {
                result = root;
            } else {
                // 當(dāng)前的根結(jié)點(diǎn)不是要找的結(jié)點(diǎn)，但是已經(jīng)找過了，所以計(jì)數(shù)器減一
                k[0]--;
            }
        }
        // 根結(jié)點(diǎn)以及根結(jié)點(diǎn)的右子結(jié)點(diǎn)都沒有找到，則找其右子樹
        if (result == null && root.right != null) {
            result = kthNodeCore(root.right, k);
        }
        return result;
    }
    public static void main(String[] args) {
        BinaryTreeNode n1 = new BinaryTreeNode(1);
        BinaryTreeNode n2 = new BinaryTreeNode(2);
        BinaryTreeNode n3 = new BinaryTreeNode(3);
        BinaryTreeNode n4 = new BinaryTreeNode(4);
        BinaryTreeNode n5 = new BinaryTreeNode(5);
        BinaryTreeNode n6 = new BinaryTreeNode(6);
        BinaryTreeNode n7 = new BinaryTreeNode(7);
        BinaryTreeNode n8 = new BinaryTreeNode(8);
        BinaryTreeNode n9 = new BinaryTreeNode(9);
        n1.left = n2;
        n1.right = n3;
        n2.left = n4;
        n2.right = n5;
        n3.left = n6;
        n3.right = n7;
        n4.left = n8;
        n4.right = n9;
        print(n1);
        System.out.println();
        for (int i = 0; i <= 10; i++) {
            System.out.printf(kthNode(n1, i) + ", ");
        }
    }
    /**
     * 中序遍歷一棵樹
     * @param root
     */
    private static void print(BinaryTreeNode root) {
        if (root != null) {
            print(root.left);
            System.out.printf("%-3d", root.val);
            print(root.right);
        }
    }
}

運(yùn)行結(jié)果

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