代码随想录算法训练营第20天|二叉树

654. 最大二叉树
思路和由中后序列构建树的思路是一样的，只不过这里不是通过后序的最后一个数来进行分割，而是通过找最大值进行分割左右子树
迭代找出左子树和右子树的的边界，返回当前节点就行了
具体代码后续补

617. 合并二叉树
这里的return node不能到最后才去return，一定在某一个条件里面马上return了

/**
 * 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 {TreeNode} root1
 * @param {TreeNode} root2
 * @return {TreeNode}
 */
 
var mergeTrees = function(root1, root2) {
    // console.log(root1,root2);
    // console.log(root1 === null)
    // return;
    //两个都为空
    if(root1 == null && root2 == null) return null;
    //root2为空
    let node = new TreeNode();
    if(root1 != null && root2 == null){
        let node = new TreeNode(root1.val);
        node.left = mergeTrees(root1.left, null);
        node.right = mergeTrees(root1.right, null);
        return node;
    }else if(root1 == null && root2 != null){
        let node = new TreeNode(root2.val);
        node.left = mergeTrees(null, root2.left);
        node.right = mergeTrees(null, root2.right);
        return node;
    }else{
        let node = new TreeNode(root1.val + root2.val);
        node.left = mergeTrees(root1.left, root2.left);
        node.right = mergeTrees(root1.right, root2.right);
        return node;
    }
};

700. 二叉搜索树中的搜索

/**
 * 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 {TreeNode} root
 * @param {number} val
 * @return {TreeNode}
 */
var searchBST = function(root, val) {
    if(!root) return null;
    if(val > root.val){
        return searchBST(root.right, val);
    }else if(val < root.val){
        return searchBST(root.left, val);
    }else return root;

};

98. 验证二叉搜索树
这道题有陷阱的，子树是BST，不代表这个树一定是BST
要清楚的特性是对于BST来说，起中序遍历的数组一定是一个升序数组

/**
 * 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 {TreeNode} root
 * @return {boolean}
 */
var isValidBST = function(root) {
    let inorder = [];
    inOrder(root, inorder);
    for(let i = 1; i < inorder.length; i++){
        if(inorder[i - 1] >= inorder[i]) return false;
    }
    return true;


};

function inOrder(root, inorder){
    if(!root) return;
    inOrder(root.left, inorder);
    inorder.push(root.val);
    inOrder(root.right, inorder);
}