654. 最大二叉樹(shù)

題目連接：https://leetcode.cn/problems/maximum-binary-tree/
思路：使用前序構(gòu)建二叉樹(shù) 區(qū)間[left, right)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private TreeNode constructMaximumBinaryTree(int[] nums, int left, int right) {
        if (left >= right) return null;
        // if (right - left == 1) return new TreeNode(nums[left]);
        int maxIndex = left;
        for (int i = left + 1; i < right; i++) {
            if (nums[i] > nums[maxIndex]) {
                maxIndex = i;
            }
        }
        int val = nums[maxIndex];
        TreeNode treeNode = new TreeNode(val);
        treeNode.left = constructMaximumBinaryTree(nums, left, maxIndex);
        treeNode.right = constructMaximumBinaryTree(nums, maxIndex + 1, right);
        return treeNode;
    }
    public TreeNode constructMaximumBinaryTree(int[] nums) {
        if (nums == null || nums.length == 0) return null;
        return constructMaximumBinaryTree(nums, 0, nums.length);
    }
}

二戈擒、 617. 合并二叉樹(shù)

題目連接：https://leetcode.cn/problems/merge-two-binary-trees/
思路一：使用前序遍歷兩個(gè)二叉樹(shù)，遇到root1 == null return root2;遇到root2 == null return root1;

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if (root1 == null) return root2;
        if (root2 == null) return root1;
        root1.val += root2.val;
        root1.left = mergeTrees(root1.left, root2.left);
        root1.right = mergeTrees(root1.right, root2.right);
        return root1;
    }
}

思路二：使用一個(gè)stack, 依次存放root2 root1, root1.val += root2.val,最后返回root1

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode mergeTrees(TreeNode root1, TreeNode root2) {
        if (root1 == null) return root2;
        if (root2 == null) return root1;
        Stack<TreeNode> stack = new Stack<>();
        stack.push(root2);
        stack.push(root1);
        while (!stack.isEmpty()){
            TreeNode treeNode1 = stack.pop();
            TreeNode treeNode2 = stack.pop();
            treeNode1.val += treeNode2.val;
            if (treeNode1.right != null && treeNode2.right != null) {
                stack.push(treeNode2.right);
                stack.push(treeNode1.right);
            } else {
                if (treeNode1.right == null) {
                    treeNode1.right = treeNode2.right;
                }
            }
            if (treeNode1.left != null && treeNode2.left != null) {
                stack.push(treeNode2.left);
                stack.push(treeNode1.left);
            } else {
                if (treeNode1.left == null) {
                    treeNode1.left = treeNode2.left;
                }
            }
        }

        return root1;
    }
}

三艰毒、 700. 二叉搜索樹(shù)中的搜索

題目連接：https://leetcode.cn/problems/search-in-a-binary-search-tree/
思路一：二叉搜索特性筐高，左子樹(shù)都比根節(jié)點(diǎn)小，右子數(shù)都比根節(jié)點(diǎn)大 使用遞歸法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
        if (root == null) return null;
        if (root.val < val) {
            return searchBST(root.right, val);
        } else if (root.val > val) {
            return searchBST(root.left, val);
        } else {
            return root;
        }
    }
}

思路二：使用迭代法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public TreeNode searchBST(TreeNode root, int val) {
        TreeNode cur = root;
        while (cur != null) {
            if (cur.val < val) {
                cur = cur.right;
            } else if (cur.val > val) {
                cur = cur.left;
            } else {
                return cur;
            }
        }
        return null;
    }
}

四丑瞧、 98. 驗(yàn)證二叉搜索樹(shù)

題目連接：https://leetcode.cn/problems/validate-binary-search-tree/
思路一：使用中序 記錄前一個(gè)節(jié)點(diǎn)凯傲，如果pre.val >= root.val 就返回false (使用遞歸中序方法)

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private TreeNode pre = null;
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        boolean isLeft = isValidBST(root.left);
        if (!isLeft) return false;
        if (pre != null && pre.val >= root.val) return false;
        pre = root;
        return isValidBST(root.right);
    }
}

思路二：使用中序迭代法

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    public boolean isValidBST(TreeNode root) {
        if (root == null) return true;
        TreeNode cur = root;
        TreeNode pre = null;
        Stack<TreeNode> stack = new Stack<>();

        while (cur != null || !stack.isEmpty()) {
            if (cur != null) {
                stack.push(cur);
                cur = cur.left;
            } else {
                TreeNode treeNode = stack.pop();
                System.out.println(treeNode.val);
                if (pre != null && pre.val >= treeNode.val) return false;
                pre = treeNode;
                cur = treeNode.right;
            }
        }
        return true;
    }
}

思路三：使用后序遍歷

/**
 * Definition for a binary tree node.
 * public class TreeNode {
 *     int val;
 *     TreeNode left;
 *     TreeNode right;
 *     TreeNode() {}
 *     TreeNode(int val) { this.val = val; }
 *     TreeNode(int val, TreeNode left, TreeNode right) {
 *         this.val = val;
 *         this.left = left;
 *         this.right = right;
 *     }
 * }
 */
class Solution {
    private boolean isValidBST(TreeNode root, long min, long max) {
        if (root == null) return true;
        if (root.val <= min || root.val >= max) return false;
        boolean isLeft = isValidBST(root.left, min, root.val);
        if (!isLeft) return false;
        return isValidBST(root.right, root.val, max);
    }
    public boolean isValidBST(TreeNode root) {
        return isValidBST(root, Long.MIN_VALUE, Long.MAX_VALUE);
    }
}